So, if you double the speed you quadruple the force of drag.

There's something else to consider. The v^2 in the equation is an artefact of the kinetic energies involved in colliding with molecules impeding your movement. When you double your speed the collisions have quadrupled in force, but in a given time interval t you have also doubled the number of molecules you are colliding with because you are moving thru them twice as fast.

That makes the total drag at any given instant proportional to v^3, total energy expenditure going from point A to point B is still proportional to v^2 (trip time is reduced if speed is increased).

Right - thus the power output of the engine has to double again, as I stated originally.

Indeed, I didn't read closely enough.