You've got quite a lot of handwaving going on, so a little more shouldn't hurt.
I recently came across this fun little question on Physics.SE: Will the Large Hadron Collider “explode” if the power is turned up too high? It, it offered a fantastic link: Protecting the LHC From Itself. It turns out that when you have 350MJ of beam energy, the energy of a speeding fright train, you have to figure out what to do when the beam does something you didn't want.
When the time comes, the beams are extracted, or dumped, into two huge cylindrical blocks. Eight meters long, one meter in diameter, and made of graphite composites encased in concrete, they are the only thing that can withstand the full power of the beam. But first the beam has to be diffused, because in its compressed form it would drill a hole tens of meters long in any material.
So as the beams pass out of the LHC, they spread out and hit the blocks in a shape that resembles a cursive “e.” The dump takes just eighty-millionths of a second, dilutes the energy of the beam by a factor of 100,000 and heats the center of the lines that make up the “e” to almost 700°C.
I point all of this out because it seems to be the current limits of what we can do with massive expensive machinery. That seems like a good starting point for your nanomachines carried by every solider on a battlefield.
The LHC's goal is to be able to focus all of that beam energy on a 65um circle. Your comments suggest 5nm is your goal, so to be fair let's convert the LHC's beam width into nanometers: 65000nm. This means the area of your wormhole is about 1/169000000th the diameter of the LHC beam.
Plugging these together, 350MJ / 169000000 = 2J. Wait a sec, that's not the units you wanted. That's the energy which passes past the LHC's focal point every time the protons pass around their racetracks. You don't want energy, you want power, so we need to factor time into this. As it turns out, the bunches of protons cross once every 25ns, or 40000000 time a second.
What does this mean? It means that if you want to transmit 2J every 25ns, you'll transmit 80MW. This means that, if you think you need to transport 80MW, the conditions of the wormhole are going to be comparable to that of the core of the LHC where we smash things together with such ferocity that we can peer into the inner fundamentals of material itself. At 80MW, you're going to want to have safeguards on the order of the LHC's massive 8 ton steel bricks which soak up the energy from its superconducting magnets. My general recommendation would be to retain a healthy safety margin away from this point. 1MW is probably much healthier.
Alternatively, since your goal is a beam weapon, just don't bother catching the beam and converting it back to energy. Just use it directly:
In 2003, two-thirds of the superconducting magnets in the Tevatron’s six-kilometer ring quenched at the same time. The beam drilled a hole in one collimator and created a 30-centimeter groove in another.