M. Werner is close, I think, but not quite there. More important than the volume of air per se, is the pressure differential between the compressions and rarefaction of the air and the surface area over which the kinetic transfer takes place. It's all about the top: the "drumhead" of the guitar. A larger top will hit more air as it rises and pull more air as it falls. A softer wood will beat the air harder by offering less resistance to the source vibration.
The trade-off happens because if the wood is too soft won't endure the stress of the vibration. It shakes itself to pieces. But if the wood is too hard it won't vibrate with sufficient amplitude. Various coatings including chitin from crusteceans permit a softer wood without losing structural integrity.
So for maximum volume you want large dimensions (particularly the top), and as soft a wood as will endure the tension.
Also, strings with greater mass will transfer more energy to the top. There's a brand called D&R (perhaps, others too) that has a "compressed" string. The make .030 gauge wrapped string by taking a .032 (or higher) and then squeezing it down to .030. It bends like a .030 'cause it's at normal tension, but the increased mass means more signal.
But beyond that are flatwound strings, which lose a lot of the upper harmonics. You do get more of the fundamental, but those upper harmonics carry most of the "volume" because higher pitches achieve greater sonic intensity (in the act of hearing) at the same physical intensity. If you have a Bass singer and a Soprano yelling at the same volume, you can hear the Soprano from farther away. That's just the way human hearing works, it's not in the physics.
The same mass argument would tend to suggest that strings with larger atomic mass would transfer more energy. But then elasticity comes in and I don't understand that part.