but the MOSFET's gate capacitance is also a factor - since the leakage current is AC, to do damage it must be large enough to charge the MOSFET gate up to its gate oxide breakdown voltage in under half a mains cycle.
Yes good point. I did kill a few 2n7000/2n7002's with my homemade lab PSU powered by a linear step-down toroidal transformer followed by a diode bridge and a filter cap. It was easy: connect the output of the PSU to the circuit assembled on a breadboard which the transistor is a part of, then touch the gate of the poor transistor intended to be used as a switch (unless protected with clamping diodes, TVS, zener etc.) -- done, it's dead. At the same time, beefier transistors, the "power" ones, with 2 nF or more gate capacitance, survived the same without issues, or at least no immediately noticeable issues.
To give an idea about what numbers we're speaking of, I measured the potential between the secondary winding (actually between any of the output terminals of the DC/DC converter that follows the transformer and rectifier) and my own body via the 10 MOhm resistance of a DMM.
It measured 9V RMS (12.7V peak) when I was floating, and 23V RMS (32.5V peak) when I was grounded (to mains PE). The latter is enough to kill a mosfet (at least a smaller one), and the former is also enough, since it's the reading under a relatively, for the current in question, low load resistance: if I measure the same with two voltmeters connected in series (~20 MOhm total resistance), then, as expected, each reads about 8V RMS, meaning 16V RMS total (or 22.6V peak), and it will continue to increase further as the load resistance is increased (up to a certain point, of course).
BTW, knowing that, we can calculate the current flowing in this scenario. Let's take 10V RMS across 10 MOhm load for simplicity, this yields 1 uA RMS.
Now, we can use it for a very rough estimation (I'm not sure we can apply this directly to the process of the gate capacitance charging, but at least it should be fine to understand the order of magnitude) of the lowest capacitance that will not be able to reach 20V during one mains voltage half-cycle.
Let's assume 50 Hz. One half-cycle is then 10 ms. Then, 1 uA (again, assuming constant current for simplicity) during 10 ms creates a charge of I*t = 1e-6 A * 10e-3 s = 1e-8 C. To have this amount of charge, the capacitor charged to 20V must have a capacity of C = Q/U = 1e-8 C / 20 V = 500 pF. This is somewhere in between smaller and bigger mosfets and agrees with what I have observed in practice.