Driving a resistive load with a square wave at a particular voltage will deliver twice as much power as driving a resistive load with a sine wave of the same peak voltage. A speaker, however, does not represent a resistive load; the current required to change a speaker's position quickly is much greater than what's required to change its position slowly. To displace air with a perfect square wave would require an infinite amount of power(*). In practice, trying to drive a speaker with a perfect square wave will result in air motion which is not a perfect square wave. While the actual observed behavior will vary depending upon the type of speaker one is trying to drive, driving a speaker with a square wave of a certain peak voltage will take a lot more power than driving it with a sine wave of the same voltage and frequency, and in general the closer the speaker is able to get to a perfect square wave, the more power will be consumed in the attempt.
(*) If air were a lossless medium, it would be possible to recover all the energy one put into the air that echos back before it is removed by something else; the power one one would have to put in would thus equal the power that other things took out. In practice, though, the closer to a square wave one tries to get, the greater the losses. Instantaneously moving air would require moving it with infinite force against a finite distance; a speaker would be unable to recapture much of that energy. Indeed, because of the speaker's electrical resistance stopping the speaker quickly would likely require pushing more energy into the speaker, rather than passively recovering energy from it.
If one's objective is to achieve the maximum perceived with a given available voltage, using a square wave or sawtooth wave would be a good way to do that. If the objective is to receive the maximum perceived loudness for a given amount of power consumed, other waveforms may be better (which waveforms are best depends upon the designs of the speaker and the driver circuit).