RS485 terminating resistor power rating

I suspect the answer is in bold in the first line of your question. The resistor is between the data lines but the differential voltage is only a few volts - the common mode voltage can be anything but won't cause an increase in potential across the termination resistors.

Figure 1. Extract from Linear's TIA/EIA-485-A Standard.

The table above says that the limit is ±5 V. Rerunning your calculation gives \$ P = \frac {V^2}{R} = \frac {5^2}{120} = 0.208\ \mathrm W \$ max.

The 0.25 W resistor should be fine.

@Transistor's table says +/-1.5V min loaded and +/-5V max unloaded so with 120 Ohms the actual power will be slightly less depending on driver ESR which tends to be around 25 Ohms for 3.3V technology and a bit more up to ~50 ohms for 5V technology , so the \$V_{OD} \$ differential will be less than 5V or 0.2W, which "may" still ok.

But how hot can the resistor get?

Abs max ratings ~ 150'C depends on p/n. with power derating after 70'C is common for SMD. or a 80'C temp rise above 70'C to result in 0W rating at 150'C

So how do you estimate the temperature rise?

Using the temperature slope from R spec, [-'C/W] the max power rating drops with rising ambient temp. This is the same as a fixed room temp and rising chip temp with Used/Max power rating. Then you need to know the inside ambient in the packaging design and ambient specs.

So if an SMD chip rated for 0.25 Wmax and 0.2W is used this results in 80% of 80'C rise or > 25+64'C = 89'C at room temp.

Generally system design criteria limits a component hot spot to 85'C @ 25'C which can still burn fingers so marginally ok.

However when 3.3V technology is used the power dissipation in the terminator is Vod_max=Vcc so Pd=9/120 W, so 3.3V technology is cooler than 5V technology with a small SMD with a lower source impedance and adequate margin.