Electronic – How many (m)W should 120 ohm CAN bus terminators have

You need to be familiar with Transmission Line Theory to understand the deeper physics in play here. That said, here's the high-level overview:

How important termination is to your system is almost exclusively determined by how long the bus wires are. Here length is determined in terms of wavelengths. If your bus is shorter than one wavelength over 10, the termination is irrelevant (practically) since there is plenty of time for the reflections introduced from an impedance mismatch to die out.

Length defined in wavelengths is a strange unit on first encounter. To convert to standard units you need to know the velocity of the wave and it's frequency. Velocity is a function of the medium it travels through and the environment surrounding the medium. Usually this can be estimated fairly well through the dielectric constant of the material and assuming free-space surrounding that medium.

Frequency is a little more interesting. For digital signals (such as those in CAN), you are concerned with the maximum frequency in the digital signal. That is well approximated by f,max = 1/(2*Tr) where Tr is the rise time (defined 30%-60% of the final voltage level, conservatively).

Why it's 120 is simply a function of the design limited by physical size. It isn't specifically important which value they picked within a broad range (for example, they could have gone with 300 Ohms). However, all devices in the network have to conform to the bus impedance, so once the CAN standard was published there can be no more debate.

Here's a reference to the publication (Thanks @MartinThompson).

Your calculations for both the resistor value and wattage are off.

For the resistor, you show the formula R = V / I, but then calculate using the entire current for all three LEDs (1 A) instead of the current for just one LED.

So it should be

$$R = \frac{V}{I} = \frac{(12 - 3.1)}{.333} = 26.7\space Ω$$

For the wattage, you are trying to use the formula P = VI, but then you substitute R for the V and use the current for all three resistors (1 A) instead of just one.

So the correct calculation would be

$$(12 - 3.1) * .333 = 2.96\space W$$

or you could use:

$$P = I^{2} R = (.333)^{2} * 26.7 = 2.96\space W$$

So you could use a 27 ohm, 5 W resistor.