Technically yes there is some heat addition. But most cooling fans are very efficient and most of that energy is converted into motion.
If you are concerned you would calculate something along these lines...
12W = 12 J/sec
cp (air) = ~1J/gK
density (air) = ~1.2kg/m^3
assuming fan flow = 1cubicmeter / minute = 1200g/min
heat input = 12J/sec=720J/min
so 720J/1200g = 0.6J/g
0.6J/g / 1J/gK => 0.6deg K temperature rise of incoming air
(multiply or divide according to the airflow of your actual fan)
This is worst case assuming all of the electrical power is converted to heat.
Fans only add heat into the system if you have a temperature in their definition. I think this is the best way to include that effect vs the other methods that you describe. They are valid, but I don't think are the best fit. Regarding how much power, it isn't likely 12w. Consider that most of it probably goes to turning the fan. Go with a worst case input temp plus a couple degs maybe for the energy that is converted to heat.
You can bound the heat that the fan dissipates. Moving some mass rate x [kg/s] of air from a dead stop to some velocity v can be converted to a power P (x*v^2/2, I think.). Then the heat dissipated will be volts*amps - P. This heat will be an upper bound, because getting the air started through a turbulent fan actually involves a lot of non-conservative work. I'd be curious whether this is useful or whether the bounds would be so conservative it doesn't really help.