I don't know what the short circuit amperage is for an 18650 battery, but an RCR123 can push 16 amps in a short-circuit -- which is to say, when the chemical reaction inside the battery is capable of generating a pressure of 4.2 volts, it is also capable of pushing electrons out the tail of the battery at a rate of 16 amps, provided nothing else is getting in the way of those electrons.
The flashlight's emitter provides enough resistance to slow down the rate of flow considerably -- but if the full 4.2v pressure were applied, it would probably still allow electrons to travel the complete circuit at a rate of 3-4 amps, which would damage the emitter in short order. (This is less of an issue with tiny lights like the Lummi Wee and Raw, because their small batteries can't maintain 4.2v for long enough to overheat the emitter.) So, it's the driver's job to reduce the pressure coming from the battery to prevent the battery from burning out the emitter.
I'll be honest, I still don't really understand how DC-to-DC voltage converters work, but the basic effect is the same as a transformer -- it outputs a lower voltage than it's given as an input, which causes the maximum current flow to substantially increase to compensate. (for the time being, disregard the fact that the converter would probably burn out if you short-circuited the outputs of the converter together -- let's just pretend we have a badass converter that can handle that much current flow.) So, now you have a circuit that will maintain 3.5v of pressure for as long as the battery can provide the necessary wattage, and the maximum current flow will decrease as the battery wears down.
But there's still the emitter to deal with. At the lower voltage coming from the driver, the emitter's resistance can slow down the current flow considerably more, so (according to your own numbers) the driver's output pressure of 3.5v can only push electrons through the emitter at a rate of 1.7-1.9 amps. But, occasionally you get an emitter with a substantially lower resistance than normal, and it allows electrons to move at a substantially higher rate, like the 2.1 amp reading you were wondering about.
The driver does burn up some of the wattage coming from the battery just to do its job of voltage-regulation, but that's a pretty predictable amount nowadays with the science of DC voltage regulation being pretty much cut-and-dried. The battery tube has a small amount of electrical resistance too, but a battery tube has about the same resistance as a car battery cable -- and those allow a current flow of several hundred amps with only 12v of pressure, so its effect on the circuit is insignificant. So the only other significant factor is the resistance of the emitter, and if you happen to get an emitter with substantially lower resistance, that means you can shove more current through it and get more light out of it. I think that's the answer you were looking for.