Can mosfets discharge power capacitors?
There are two parameters in particular to look at.  You have to make sure you don't exceed the maximum current, and the fast energy dump can't heat the die too much.

The first is obvious.  Take the maximum capacitor voltage, divide it by the maximum allowed FET current, and that is the total series resistance needed.  In theory, the cap's ESR and the FET's on-resistance contribute to the total series resistance.  However, those two are usually specified as maximums that will not be exceeded, so you can't count on minimum values.  The safe thing to do is to size the resistor as if it were there only series resistance.

Once you know the current is limited, you have to make sure the FET doesn't get too hot.  Here you probably need to do some guessing and derating.  The datasheet will tell you the maximum continuous dissipation the FET can handle for whatever conditions of cooling and heat-sinking you set up.  However, it doesn't usually tell you how much energy it takes just to get to a particular temperature.  You can exceed maximum continuous dissipation for a short term when starting from cold, but it's often not specified how much more dissipation for how long.  Use common sense and be conservative.

One strategy is to have the resistor dissipate most of the cap's energy.  In this case you use the FET's maximum guaranteed on resistance.  The fraction of the total energy dissipated by each series element is its resistance divided by the total.  For example, if the deliberate resistance is 500 m&Omega; and the FET's maximum on resistance is 25 m&Omega;, then the FET is guaranteed to not dissipate more than (25 m&Omega;)/(525 m&Omega;) = 4.8% if the cap's energy.

A larger series resistance slows the cap discharge time, and shifts more of the total dissipation to the resistor.