What is the Half-Life of a Large Chunk of Fissile Material?

As the other answers point out, half-life is a property of the isotope rather than the chunk. It can be defined as the time when there is 50% chance of any given nuclei having decayed spontaneously: there is no reference to any outside effects. The spontaneous decay of nuclei can be affected by their surroundings and obviously if you irradiate them a lot of things can happen.

The term "critical mass" is also a problem. If you have 52 kg of uranium in the form of a long wire it will not spontaneously explode. The neutrons released from spontaneous fission will mostly shoot away into the air. If you roll it up, things might get hotter. Alex Wellerstein has a very good post about why the term "critical mass" confuses things. A better term would be "critical condition": a condition where the spontaneous emission of neutrons will cause a chain reaction. This depends not just on mass, but the shape, presence of neutron reflectors, and other factors.

So, how do you actually calculate things? Basically you need to calculate the rate of emission of neutrons, and how these neutrons then diffuse in the chunk and cause more emission. This gets complicated quickly: the rate of stimulated emission depends on a cross-section that depends on neutron energy (this cross section can be measured but is not a simple function), and produce neutrons with random energies according to some probability distribution (another thing to measure). These neutrons then diffuse through the material as they bounce off other atoms, and one has to solve the diffusion equation for them (a fairly standard boundary value problem). Things of course get messier since they do produce more neutrons, so the actual calculation will be a diffusion equation with multiplication.

This is still a calculation for a steady-state situation rather than a critical situation where things grow over time; that adds extra mathematical wrinkles that I suspect are not readily available in the open literature. In any case, the total calculation is basically a model of the rate of neutron generation and absorption given the conditions, and we learn the critical condition by seeing where it shoots off to infinity. In practice the calculations are done numerically on a computer rather than analytically because of all the above complications but there are likely approximations one could run with pen and paper to get a rough sense of what is going on (after all, the Manhattan project did it successfully, although they sometimes used analogue computers like FERMIAC).

A fissionable material undergoing a chain reaction doesn't have a half-life. A half-life is a parameter that describes a tailing exponential curve. That's not what we have in the case of a chain reaction.

Your definition of "fissile" only applies after they have acheived critical mass. And critical mass is a slippery number; if 52kg is called out as the critical mass (at STP), that probably presumes spherical arrangement; a cube would be subcritical.

And even if it was critical, that would be delayed critical - that is, some of the neutrons that keep it critical actually occur from fission products which have already been split, decaying further by neutron release, and subject to their own half-lives. This level of criticality will not yield an appreciable explosion.

That will result in an excursion, an exponential increase in power at human observable time scales (seconds), until the increase in power causes its own Rapid Unscheduled Disassembly.

Which is what happened at Chornobyl, except, the only reason Chornobyl was so unbelievably toxic is that this reactor had been running for quite some time and had old fuel with lots of fission and decay products like Cs137 in it, and that's what went kablooey all over Europe. If the Chornobyl excursion had happened with all brand new virgin reactor fuel that had never been subject to fission except for this excursion, the only radioactive fission products would've been from those few seconds of excursion -- not enough for anyone in Pripyat to notice, let alone Sweden. It wouldn't have even been detected by the west, and would've been a radiological nothingburger.

And even if there was prompt criticality - enough past critical mass that the fission could be sustained exclusively with U235 splits alone -- still, this would only result in a RUD.

Actually holding a critical mass together long enough to make Sodom & Gomorrah amounts of power is actually a hard problem. You're not going to do it by accident. And even then, it isn't a question of "long enough" because the time scale we're talking about is in shakes.

So the answer to your question is "the half-life doesn't change", because it is impossible for the criticality to remain assembled long enough to matter.