jferg:
One other thing: Is the available capacity of the 5 volt supply higher when connected to a 12 volt supply than with a the typical USB connection? It seems like it would be, but how to quantify? Also, if my stupid questions might be answered by reading existing text, where should I have looked?
There are two limiters of current: The power source (wall wart, or 12V jack, or whatever,) and 5V regulator.
If your 12V supply is a marine/car battery or similar, then the "wall wart" is not a limitation here. However, with a linear regulator, you cannot get more current out of the 5V than you draw from the jack.
Second, the regulator will overheat if you draw too much, and may have current limiting to prevent that. Typically, you'll get between 500 mA and 1500 mA out of a TO-220 device, depending on the specifics. You may also want a heatsink to actually get to the full specified current. For example, getting the specified 950 mA max current out of a 1117V50 LDO regulator when powered by more than about 7V requires a heat sink screwed onto it.
You can do all this math by reading the data sheet for the regulator you're using. The things you care about are max current limit, internal resistance/dissipation, and heat conductivity to ambient. The resistance of the regulator is simply how it regulates power. If you get 12V in, and 5V out, at 0.9 A, it has to dissipate (12-5)*0.9 == 6.3 Watts. With an un-heat-sunk thermal resistance of something like 80 C per W (making this up), you'd heat up to 500 C(!), and go up in flames. With a 15 C per W heatsink, you instead heat up by about 100C, which at 20C ambient is (just) within spec.
If you want to get better efficiency, and/or draw more current at lower voltage than you put in at high voltage, you need a switching regulator/converter.