No switch is perfect. Switch selection is based on the load that you are going to handle. No load has zero inductance or zero capacitance. Hence there will be surges and kick backs. It should be handled right - if not the circuit will fail under certain inputs.
Now-a-days you could easily find FETs with very low RDS ON (even 2.0 milli Ohms is not uncommon. Typically automotive industry pays pittance for those parts - but open market pricing could be expensive. I have never tried get those outside, but I guess even if it is triple it is still within your target). But once the RDS on is low, if you ever drive it in linear mode, it could easily smoke.
You can easily use a MOSFET if the Fan specifies a weak internal pull-up and a 2N7000 fits a 12V low current application.
Whether you should worry about the fan supplying more than officially specified, is another thing. If you have shorted the pin to ground for your measurement and 0.68mA came out, that's so incredibly well within the reach of a 2N7000 that you shouldn't worry about the FET too much. You're welcome to worry about the Fan itself, if that helps you in any way, but a quick test with an arbitrary PWM signal will show if it is broken, or just built different to (EDIT:) the official prescribed spec.
All in all, I say, go for it. If you are really doubtful, you can limit some FET damage by adding a resistor in the drain path, like so:
simulate this circuit – Schematic created using CircuitLab
The 2N7000 has an absolute worst case drain-source resistance of 13.5Ohm (the one Fairchild makes, at least) and I expect in your use-case that it will not exceed 3Ohm even when hot.
(EDIT2:
Sunday made my maths go wobbly, changed the voltage drop and the conclusion to it in the following block:)
So at the very worst the resistor and the FET will share the risk, normally the resistor will catch the brunt of it. At 5mA maximum the resistor will only waste 0.05V, which is negligible, you can even increase it to 20 or 30 ohm and it should still be interpreted as low. In the worst kind of failure (PWM shorted to 12V in the fan) the 10Ohm will effectively limit the current through the FET to about 1A (if the FET also is at least 2 ohm then), which is an acceptable very short-term limit. 30Ohm would probably keep the FET fully safe until the resistor itself burns through.
9999 out of 10000 times the resistor is completely redundant and pointless. But it is an option if it helps you sleep better (2 cents for good sleep, I call a steal)
Best Answer
That is not a suitable transistor for this application since you want to drive the base with 3.3V. The 60V rating is also overkill.
A much better fit for this application is the IRLML2502. That is specified for lower gate voltage. It can only handle 20V, but that's well above your 12V spec.
The body diode is not in the right place to serve as a load flyback diode. Think of the load as a inductor, and you'll see that it will try to make a high voltage when turned off, not a low voltage. You still need a reverse diode accross the load (the fan in this case). Make sure the diode is rated for at least the same current as the fan. A regular silicon diode will be fine as long as you turn the fan only on or off, and don't try to turn it on shortly after having turned it off. If you plan to do PWM control, then use a Schottky diode since those have essentially zero reverse recovery time for this purpose. It will also be slightly more efficient. At 12V it will be no problem finding a suitable Schottky diode.