Electronic – Why would a computer power supply output 5.2V on the 5V rail

One way to "bullet proof" your power supply is to add transient voltage suppresors (TVS). See this application note for instance: http://www.vishay.com/docs/88490/tvs.pdf‎

Also check the fuse configuration of the ATmega32. Some of them may help to make your system more robust. In particular, check the Brown-out detection module and the start-up time (SUT0 and SUT1 fuses).

Just because the continuity test function of your meter does not detect a short does not mean there is an unintended path for excessive current between the supply rails. There are many potential faults that could exist but won't be detected with a meter like that, but here's one: a forward-biased diode between the rails.

Most meters, except the very cheapest of them, test continuity with less than 0.6V precisely so that they can't forward-bias a silicon P-N junction. Thus, they will not detect a diode as "continuity" regardless of which way you place the diode with respect to the meter. However, to any power supply significantly more than 0.6V, the diode might as well be a dead short, if it's forward-biased.

If your circuit is simple, the easiest way to troubleshoot is usually to inspect it very carefully, looking for errors. If that fails, you can connect the power supply and leave it in CC mode, with the current set low enough that you aren't burning anything. Then, with a good, precise voltmeter, start measuring the voltage drop between points on your board. Where there is a voltage drop, there is high current (remember, \$E=IR\$), and thus you know the fault is somewhere between those two points. Be methodical about it and you should be able to find the fault by process of elimination.