All silicon is photosensitive to some extent. The power supply will be one of the few chips which is primarily "analog"; in particular it will have a bandgap voltage reference. That's exactly the sort of thing to be badly affected by a charge pulse from the flash. Glitching the power supply will then destabilise the digital logic it's powering.
The real question is how the light manages to get through the epoxy casing.
Yes, but if you were tasked with designing the epoxy that was going to be used in every IC everywhere, I would hope that you would go through the extra effort of adding dyes that were specifically opaque to UV and IR.
All chips are photosensitive and a xenon flash produces light from UV through to IR, so my guess would be that the plastic used for the casing on that chip is not entirely opaque to something at the top or bottom end of the visible spectrum, which is why most bright lights don't trigger it. At a bet, I'd say the top end, as otherwise radiant heaters would also probably kill them. Would be fun to play around with some filters and a flashgun and narrow down the frequency.
I think it's more likely to be the other way round, a lot of plastics are semi-transparent to IR, all block UV extremely strongly (as you start ionising electrons).
For instance, black PVC electrician's tape is entirely transparent to near-infrared, I use this to cover Xenon flashbulbs & still have an infrared-sensitive trigger fire from them.
That was my first guess as well, the only reason I discounted it was that then I would expect the effect to happen more often. I could be entirely wrong there however.
edit - ars, I'm out of replies, but thanks for that. I think you are probably right.
