The formula for escape velocity is
$$v_e = \sqrt{\frac{2Gm}{r}}$$
$v_e$ is the escape velocity.
$G$ is the gravitational constant, approximately equal to $6.67×10^{−11} m^3\cdot {kg}^{−1}\cdot s^{−2}$.
$m$ is the mass of the object from which we are trying to escape (the black hole).
$r$ is the distance between that object and the object that is trying to escape.
The formula for the magnitude of gravitational acceleration is
$$g = \frac{Gm}{r^2}$$
$g$ is the acceleration due to gravity.
Now, we know $G$ (a constant), $v_e$ (greater than the speed of light), and $r$ (either 10 or 250 meters: given in your question). We do not know $g$, although we know that it is at most fifteen times the Earth's acceleration due to gravity at sea level. We do not know $m$.
So let's rewrite the first equation in terms of $m$.
$$m = \frac{{v_e}^2r}{2G}$$
Now we substitute that into the second equation.
$$g = \frac{{v_e}^2}{2r}$$
Now, substitute in the speed of light and the distance. We'll use 10 meters as the distance.
$$g = 4.5\cdot 10^{15} \frac{m}{s^2}$$
That's roughly $4.5\cdot 10^{14}g_E$, where $g_E$ is the acceleration due to gravity at sea level on the Earth. That's much higher than fifteen.
We can calculate things the other way. A 15g acceleration at the event horizon of a black hole implies a black hole that is less than $3\cdot 10^{-13}$ meters in size. This is tiny. For comparison, a proton is about $10^{-15}$ meters. So this is bigger than a nucleus but a lot smaller than an atom.
I have ignored relativistic effects, but I don't think they would matter much. We start in the same inertial frame. There might be a black hole, but it would be smaller than an atom.
You could improve on this by making the initial acceleration dependent on distance as well. For example, if he could make an acceleration equal to fifteen g at a distance of ten meters from the center of attraction, then it would be a lot larger at one meter and larger still at one millimeter.
Part of the confusion may be that a black hole still has gravitational attraction outside its event horizon. The event horizon is the place where the black hole won't release light. But you'd have to moving close to the speed of light to escape if you came close to the event horizon.
Your fifteen g of acceleration is confusing in that it doesn't mention the distance to the center of attraction. The presumption is that it is being created at the accelerated object. But gravity doesn't work that way. It accelerates towards a center of mass somewhere.
It might make more sense to think of him as always creating a black hole. The stronger he gets, the farther away that he can pull an object. The black hole might get visible at sufficient strength.
For example, consider if he can create a fifteen g acceleration at 250 meters from the point of attraction.
Let 250 meters be $d$ and solve the second equation for $m$.
$$m = \frac{gd^2}{G}$$
Now solve the first equation for $r$.
$$r = \frac{2Gm}{v_e^2}$$
Substitute
$$r = \frac{2gd^2}{c^2}$$
$c$ is the velocity of light.
This gives us about a $10^-10$ meters black hole. That's similar to the size of an atom, so still far below visibility.
TL;DR: fifteen g is not enough to make a visible black hole, even operating at a distance of 250 meters.
Gin that equation is not Earth's gravity acceleration. It's the gravitation constant (6.67408 × 10-11 m3 kg-1 s-2).Fin that equation is equal tomg, wheremis the mass that is not the Earth's mass. So $g_E = \frac{G*m_E}{r_E^2} $ $\endgroup$