Do the math - Voltage (12V) total series resistance (max 50K + typically 160) therefore Current and Power = 2 fifths of Fcuk all
eh? Let's assume the fan is a purely resistive load (it's not, but it'll do for argument's sake). If the label says 12V 300mA, it's a 40ohm load. If you use a 500ohm, 0.5W pot to slow it down as SJW has, the responses are shown in the chart below.
Note that the system current drops off sharply, which is what I mean when I talk about non-linearity of control. If we assume the fan speed to be roughly proportional to system current, this means it'll be very twitchy at the top end, with a long tail - it takes a small turn to cut the speed to very low, where it will stay over most of the pot's range.
Note also that the power dissipated in the pot peaks at about 8%FSD, with 0.9W. This nearly doubles the rated power of the pot, but the pot's rating is across the entire range. If you're using 8% of it's length, you've only got 8% of its power rating, or 0.04W. You're therefore overpowering the pot by 22.5 times. The power is only within rating above about 340ohms, in which range the system current is pretty well constant.
Since the pot is probably well underrated, this isn't likely to cause immediate and catastrophic failure, but I'd say it's a fair way north of 2 fifths of Fcuk all.
Is there any other math you'd like to suggest I do, Screwy?