Best way to drop 12V fan to 7-9V?
At 12V 0.2 amps, you have 60 ohms equivalent motor resistance at speed (Ohm's law).

Using voltage division:

$$V_{out} = V_{in} \frac{R_1}{R_1+R_2}  $$
$$\frac{V_{out}}{V_{in} \cdot R_1} = \frac{1}{R_1+R_2}  $$
$$R_1 + R_2 = \frac{V_{in} \cdot R_1}{V_{out}}  $$
$$R_2 = \frac{V_{in} \cdot R_1}{V_{out}} - R_1  $$
$$R_2 = R_1 \cdot (\frac{V_{in}}{V_{out}}-1)  $$
$$R_2 = 60(\frac{12}{8} - 1) = 60 \cdot 0.5 = 30 \Omega$$

$$30 \Omega \cdot (0.2 A)^2 = 1.2 W$$

A 1.5 or 2 watt resistor will suffice if it is 30 ohms. You can go bigger if you like, but it won't do much more.

Make sure your fan still spins well though. If it doesn't you could be drawing more current than you want. In that case, up the voltage some. I would find a high power Potentiometer to test this out with. Or perhaps get a couple different resistance values to try out what's the optimal amount of fan cooling vs sound.