I'd say you'll have to calculate the speed it could actually achieve by v max = (Pavailable/D)1/3, because it can only accelerate as long Pavailable > Prequired.
Using your figures I get v max ~ 66 m/s, assuming it has no drag from still being in the water.
The next step I think should be to figure out how much lift is generated at that speed [I'll see if I can figure out that JavaFoil-thingy]
and see if it's enough to get it of the ground.
Also, I would assume the weight to be something to between 150000kg (MTOW - one Sherman tank) and 90000kg (MTOW - 750 troops) because I think the using the projected payload is a little ambitious for a first prototype.
Now tear it apart while I go back to (not) studying physics . . .
You've got good points here! Too tired to redo the calculations now, but I'll probably do some revising the day after tomorrow. Since the velocity to the third power determines the required power, small differences in this could make a large difference. Also, I've been searching for the power these engines deliver, and it could be it's about 250W/engine more than I used here.
About the weight: it's not yet used in this calculation, since it's only used for determining climb rate, not power required.