Your planet probably can not stay together if if rotates as fast as you request.
I am unable to calculate the correct ratation rate to maintain a specific shape for a panet of a specified mass, nor the correct rotation rate to reduce the surface gravity at the equator to 1 g. So of course I cannot design a planet where those two rotation rates would be the same rate, thus making it possible for such a planet to exist. Perhaps someone else can do it for you.
Howevery, I have been able to calculate that it is probably impossible for a planet to stay together if it had a half hour long rotation period. Any large and oblate planet would have an equatorial rotatin speed greater than its escape velocity if it rotated that fast.
You didn't say that the planet had to be habitable for humans. But you did require that the planet had to: 1) have a surface gravity at the equator which was survivable for humans, and 2) be as large as possible, - Jupiter mass was mentioned.
Okay, I now notice that the question requires 1 g of surface gravity at the equator. So I don't have to consider what is the highest surface gravity humans can tolerate.
The mass of Jupiter is about 317.8 times the mass of Earth, so we should make the mass of the planet 300 times the mass of Earth for simplicity.
Assume for the moment that the planet somehow has an average surface density equal to that of the Earth, 5.514 grams per cubic centimeter. Assume that it is shaped like a very flat cylinder. Assume that its thickness is approximately equal to the diameter of Earth, 12,742 kilometers.
To have 300 times the mass of Earth and the same average density as Earth, the planet would have to have 300 times the volume of Earth.
A sphere with the radius of Earth has a volume of 4.18879 cubic Earth radii.
volume of sphere calculator
A cylinder with the radius of Earth and a height of 2 Earth radii has a volume of about 6.28 cubic Earth radii, about 1.499 times the volume of a sphere with the radius of Earth.
https://www.omnicalculator.com/math/cylinder-volume
300 divided by 1.499 is about 200.13342. The square root of 200.13342 is about 14.146581. So a flat cylinder with a radius of 14.146581 Earth radii and a height of 2 Earth radii (I Earth diameter) should have a volume of 300 Earth volumes.
The cylinder volume calculator gives such a cylinder a volume of 1,257.43 cubic Earth radii, or 300.18931 times Earth's volume. So such a cylinder would be 2 Earth radii or 12,742 kilometers "high" and have a radius of 90,127.867 kilometers and diameter of 180,255.73 kilometers.
The average radius of the Earth is 6,371 kilometers, but it varies between 6,356.752 kilometers at the poles to 6,378.137 kilometers at the equator, a difference of 21.385 kilometers. The Earth is oblate because it rotates, and the faster movement at he equator lifts up the surface of the Earth and reduces the surface gravity there a tiny bit.
The pancake shaped world would have to rotate rapidly to maintain its shape. Othewise its gravity would pull it into a sphere with a radius of about 6.69573724 Earth radii.
But assume that it wasn't rotating. What would be the surface gravity at the edge of fhe cylinder - equivaent to the equator of a pancake shaped planet - be?
According to my rough calculations, the surface gravity of a world with 300 times the mass of Earth at a distance of 14.146581 Earth radii would be about 300 divided by 200.12575, the square of 14.146581, and thus about 1.4990574 g which is about 14.705 meters per second per second.
Possibly some people wouldn't mind walking around in 1.499 g. though possibly the weight of space suits or other environmental protection would be too much for them. But maybe it turns out that nobody can stand to walking around for long in more than 1.25 g.
Anyway, the problem is to get the surface gravity down to 1 g at the rim of the cylinder that corresponds to the equator of the pancake planet.
I don't know how to calculate the proper rotational speed to bring the surface gravity of the pancake planet down to 1 g at the equator.
But it should be possible to calculate a maximum possible ratational speed of such a pancake planet.
The escape velocity at the rim of the cylinder roughly corresponding to the pancake planet equator can be calculated. The planet would have 300 times the mass of Earth, so the escape velocity at a distance of 14.146581 Earth radii would be 51.51 kilometers per second.
https://www.omnicalculator.com/physics/escape-velocity
Since the pancake planet would 14.146581 times the radius of Earth it would also have 14.146581 times the circumference of Earth. So the cylinder would have a circumference of 566,924.47 kilometers. The traveling at 51.51 kilometers per secondit would take 11,006.105 seconds, or 3.057 hours, to travel 566,924.47 kilometers. So if the pancake planet rotated faster than once every 3.057 hours material at its equator would be taravelling faster than escape velocity and wuld escape into outer space.
The question asks for a planet rotating about once every half hour, which is about 6 times as fast as the maximum speed that a planet with the above mass and dimensions could rotate at before starting to break up.
So this example of a pancake planet has to be rotating very fast to maintain its oblate shape, but also has to rotate at the proper speed to have a surface gravity of only 1 g at its equator, and also has to rotate less than once every 3.057 hours to avoid breaking up. And I don't know if the proper speed to maintain its shape would be equal to the proper speed to have a surface gravity of 1 g on the equator.
The question asked how large such a planet could get in mass and radius.
And the easy - though possibly not correct - answer is that a pancake shaped planet could get as massive as any planet can get. The dividing line in mass between the most massive planets and the least massive brown dwarfs is believed to be about 13 times the mass of Jupiter, and thus about 4,131.4 times the mass of Earth. Brown dwarfs are neither planets nor stars, but rather intermediate types of objects.
So we can assume that a pancake shaped planet might possibly have as much as 4,000 times the mass of Earth. How it would form is another question. Explorers from Earth might believe that such a massive planet without much light elements was impossible to form naturally and theorize that a very advanced civilization actually built that planet, perhaps while experimenting to build an Alderson disc. But that is another story.
And this time I will make the planet several times as "thick" in my crude cylinderical model. I will try making it 10 Earth radii or 5 Earth diameters thick.
A cylinder with a radius of 1 Earth radius and a height of 10 Earth radii would have a volume of 320.44 cubic Earth radii or 76.499418 times the volume of Earth. A world with the density of Earth and 4,000 times the mass would need a volume 4,000 times the volume of Earth.
A clyindrical planet with a polar diameter of 10 Earth radii (63,710 kilomeers)) and an equatorial radius of 23.09401 Earth radii (147,131.93 kilometers) and diameter of 46.18802 Earth radii (294,263.86 kilometers) would have 4,000 times the volume of Earth.
Such a planet would have a surface gravity of about 7.5000046 g at the equator, and would have to rotate rapidly to reduce it to 1 g at the equator. And it would also have to spin fast enough to have such an oblate shape.
It would have an escape velocity of 147.2 kilometers per second at the equator.
https://www.omnicalculator.com/physics/escape-velocity
With a radius of 147,131.93 kilometers it would have a circumference of about 924,456.39 kilometers at the equator. So a full rotation of the planet at 147.2 kilometers per second, escape velocity, would take about 6,280.2743 seconds, or 1.74452 hours. That 3.489 times as long as the half hour per rotation requested in the OP.
Let's try to see how fast perfectly spherical planets with 300 times the mass of Earth and 4,000 times the mass of Earth could rotate without starting to lose mass - ignoring of course that they would become oblate if they rotated even a small fraction as fast as that.
A perfectly spherical planet with a radius of 6.69573724 Earth radii would have a volume 300 times that of Earth. If it had the same average density as Earth it would have 300 times the mass of Earth.
And it would have an escape velocity of 74.87 kilometers per second. With a radius of 6.69573724 Earth radii it would have a circumference of about 268,031.29 kilometers. Turning at 74.87 kilometers per second, it would take 3,579.9557 seconds, or 0.9944321 hours to rotate, about 1.9888 times as long as the requested half hour rotation.
If a perfectly spherical planet has the same average density as Earth it should have about 4,000 times the mass of Earth if it has 4,000 times the volume of Earth. Since the cube root of 3,999.9995 is 15.87401, a spherical planet with 15.8741 times the radius of Earth and the same average density a s Earthwould have a mass of 3,999.9995 Earths.
Such a planet would have an escape velocity of 17.756 kilometers per second, With a radius 15.87401 times that of Earth, it would have a circumference of 636,151.22 kilometers. Rotating at 17.756 kilometers per second it would take 35,827.394 seconds, or 9.9520 hours, for a full rotation, about 19.9 times as long as the requested rotation time.
My rough calculations show that it might not be possible to design a planet that could rotate in about half an hour witoout breaking up.
And I think that designing a planet where the right rotation rate to reduce equatorial gravity to one g was also the right rotation rate to maintain the shape of the planet would be hard enough to do.
