I don't have a very satisfactory description of the microscopic picture, but let me share my thoughts.

The Pauli exclusion doesn't quite say that fermions can't be squeezed together in space. It says that two fermions can't share the same quantum state (spin included). A black hole has an enormous amount of entropy (proportional to its area, from the famous Bekenstein-Hawking formula $S = \frac{A}{4}$) and hence, its state count is $\sim e^A$.

Now, this might not seem like a big deal since usual matter has entropy proportional to volume. However, volume of such collections is also proportional to the mass. This means that a counting of the number of states goes as $e^M$

For a black hole, it's Schwarzschild radius is proportional to the mass, hence $A \sim M^2$. So, the number of states scales as $e^{M^2}$ *which is much much more than ordinary matter*, especially if the mass is "not small". So there seem to be a *lot* of quantum states into which one can shove the fermions.

So it seems like the fermions should have an easier time in a black hole than in (say) a neutron star.

This post imported from StackExchange Physics at 2014-04-24 02:34 (UCT), posted by SE-user Siva