Skyco and Kenny are right, Watts are Watts regardless of the source. Power in Watts = current X voltage. Or P= I x E(P over IE). If you know any 2 values you can solve for the third by multiplying or dividing as appropriate. 400 Watts divided by 120V = 3.3A(P divided by E = I). That same 400W being drawn from a battery at 13V = 30A, plus a bit more for the conversion losses in the inverter.
Inductive loads such as drills, saws and refrigerator compressors can pull 2-3 times their normal running current during startup, so to feed a drill or saw with a 400W static load, may require as large as a 1000-1200W inverter. Most inductive loads(motors) also get most of their energy from the peak of the AC sinewave. Modified sinewave inverters approximate a sine wave with a series of steps and are typically a little flat topped. This deprives inductive loads of that energy so they will draw a bit more current to make up for it. Microwave ovens don't liked flat topped pulses for this reason and cook times can be much longer as they can't deliver as much energy to the magnetron with out a good ac pulse shape. Switching power supplies in computers also work a little harder on a mod AC sinewave. That being said, mod sinewave inverters will run most everything, just not as efficiently as a true AC sinewave.
As mentioned, this energy will need to come from your alternator on the tractor. Most tractor power systems are pretty marginal in my experience. Your alternator might not be adequate to feed a 30+ amp load in addition to charging the battery and running the instrunments. Short duration use could be covered by the battery, but the alternator will still try to make up for these losses while they are occuring. A 25-30A alternator might cook itself trying to feed even short duration 50A loads on the electrical system...