Your TA is right that energy density alone does not trigger black hole formation. Consider a ball that's sitting still. Now speed up and look at the ball again. It will have gained (kinetic) energy. Relativistically, you can make the ball's energy density arbitrarily large by moving sufficiently near the speed of light. But the ball hasn't done anything in this process. It's you that has been changing speed. The notion of a black hole is not observer-dependent, so energy density alone cannot make a black hole form.
That said, there are senses in which electromagnetic fields can form black holes in general relativity. Two colliding electromagnetic plane waves (which are necessarily gravitational plane waves as well) can do this. The field strength for a single plane wave can be arbitrarily large, however. Any "limits" are highly dependent on the specifics of the configuration.
The maximum force / power are obtained in the paper as:
$$\textrm{Force} \le \frac{c^4}{4G}\;\;\;\;\;\;\;\;\;\textrm{Power} \le \frac{c^5}{4G}$$
These are just 1/4 the Planck force or Planck mass one would obtain from the Planck length, mass, and time.
First noticeable weirdness is that he keeps talking about momentum and energy in high gravity situations when these concepts are not very well defined in the context of GR. I guess that's okay in that he's deriving GR rather than making a statement about GR. Later he explains that force and power are highly observer dependent.
Page 3 has another weirdness, could be some sort of typo or oversight by the author:
As a result, a maximum force
(respectively power) value in nature
implies the following statement: one
imagines a physical surface and
completely covers it with observers;
then the integral of all momentum
(respectively energy) values flowing
through that surface per time,
measured by all those observers, never
exceeds the maximum value.
The above is weird in that the maximum force should depend on the size of the surface. That is, a surface of twice the size should allow twice the force. Maybe this is incorrect intuition on my side, but it could at least use a better description.
On the same page, he notes that force and power are parts of a 4-vector in relativity. Kind of cool. I never thought about that. Might be nice to see a reference.
Reading further, he makes it clear that the "maximum forces" he's talking about are on the event horizons of black holes. That seems reasonable. In the Schwarzschild coordinates objects freeze on the event horizon but in the coordinates I prefer, Gullstrand-Painleve (GP), objects fall through and one can compute an acceleration at the event horizon. The amusing thing is that I wrote the definitive paper on this subject:
Int.J.Mod.Phys.D18:2289-2294 (2009) Carl Brannen, The force of gravity in Schwarzschild and Gullstrand-Painlevé coordinates
http://arxiv.org/abs/0907.0660
If you look at equation (11) of the above, the acceleration is zero at the event horizon for stationary observers in Schwarzschild or Gullstrand-Painleve coordinates. For the Schwarzschild case this is intuitively obvious. That's part of why particles get stuck on the event horizon. I think his force doesn't make sense.
Ah hah! He has some counterarguments to critiques of his theory. The first one is the "mountatin example". In it, he writes: Nuclei are quantum particles and have an indeterminacy in their position; this indeterminacy is essentially the nucleus–nucleus distance. This is clearly untrue; the electron wave functions define the nucleus-nucleus distance and this is far greater than the size of the nucleus wave function. One's measured in barns, the other's measured in Angstroms.
The counter-argument for "multiple neutron stars" fails to take into account how special relativity deals with simultaneous events. He's claiming that (in a given rest frame) it's impossible to have simultaneous events that are widely separated. This is a confusion of spacelike separation and simultaneity. With his use it eliminates any set of observers to be simultaneous but it seems that was an intrinsic assumption of his argument.
So my verdict is that it's junk, but better junk than most in that I didn't see any violations of units.
He uses arguments that all rely on the Planck units to argue that the Planck units are fundamental. For example, the diameter of a black hole has a close relationship between the mass of the black hole and the gravitational constant. So when you consider the time light takes to cross this distance, of course you're going to get more Planck constant related things. There's plenty of opportunities to get the right constants as there's an infinite number of ways of choosing arbitrary lengths and distances from the geometry of a black hole. No surprises.
Best Answer
The answer is NO. There is no energy density limit (for all three questions).
The easiest way to see this is that the energy density is just the $T^{00}$ component of the stress energy tensor. The solution in GR depends on the full stress energy tensor, so it is not enough to just talk about the energy density. Furthermore, because the energy density is just a component of a tensor, it is a coordinate system dependent quantity. So starting from a solution that doesn't become a blackhole, and has some energy somewhere, we can always choose the coordinate system to make the energy density arbitrarily large.
More clearly stated: Local Lorentz symmetry alone is enough to show that the energy density is not limited in GR. And furthermore since there exist non-zero energy solutions that don't become blackholes, this also answers your second question.
To make the answer to the third question more clear, let's discuss an exact solution. Consider the Robertson-Walker solution with a perfect fluid. Here's an example stress energy tensor for a perfect fluid in the comoving frame:
$T^{ab} =\left( \begin{matrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{matrix} \right)$
Now if we change to a different coordinate system, using the coordinate transformation: $\Lambda^{\mu}{}_{\nu} =\left( \begin{matrix} \gamma &-\beta \gamma & 0 & 0 \\ -\beta \gamma&\gamma&0&0\\ 0&0&1&0\\ 0&0&0&1\\ \end{matrix} \right)$
We see the energy density will transform as: $\rho' = \gamma^2 \rho + p \beta^2 \gamma^2 = \gamma^2 (\rho + p \beta^2)$
So not only can the energy density be arbitrarily large, but even over a finite volume.