That's inaccurate, that's not really how it works. Draining the battery as fast as it could be would mean short-circuiting it (which would make it catch fire in case of Li-Ion cells) The Mi heater is not a short-circuit, it has some resistance, albeit a small one (sub-ohm) but still a finite value.
This is that resistance (which happens to vary slightly with temperature but that's not the point) that determines the current that will flow through the circuit for a given voltage. It will never draw from the cell more than this maximum current but it could already exceed what the cell can continuously deliver in a safe way (the so-called CDR of the cell, constant discharge rate) if ever you were using something else than the recommended cell models.
All RBT heaters have slightly different resistance values so it's not possible to give a precise answer without measuring said value. The peak power will only be outputted when the cell is fully charged (4.2V) but as soon as you hit the trigger the cell voltage will sag (due do its internal resistance) and thus the power will only decrease from that point on and be always less than this theoretical peak value. Then the more the cell discharges, the lower its voltage, and the lower the power output.
I did some measurements on my beta Mi1 a long while ago, but it's lost somewhere in this long thread I believe .
If you want to visualize that more easily you can use this free online tool:
http://www.steam-engine.org/ohm
Keep the resistance locked (default "lock" buttons settings on the right), enter something like 0.40 ohm in the resistance field, that would represent your heater (not the real value but not that far away) Then enter 4.20V for the voltage. You'll see the current (A) and power (W) for that max theoretical fully charged value. Then use the down arrow to lower the voltage to simulate your cell discharging and watch how the A and W figures go down.