1) Typically yes, but USB hosts are sometimes very loosely implemented. I've got a cheap Sweex USB hub which doesn't do anything when I short circuit the +5V USB power..
2) Depends on the headphones and the amplifier. You can probably figure out how much it could take to look at the maximum power, and divide it by the 1.5Volts. It's probably a maximum with maximum decibels.
I don't know the exact limits of a AA battery, but if you got a 2200mAh battery, I would say they should be able to deliver 2.2Amps (1C discharge rate).
3&4) A LM317T is a simple solution. It dissipates the 'left over' volts into heat. So if the input to output difference is 3,5V at 100mA, it will dissipate 100mA*3.5V=350mW.
Note that if you take 100mA at 1.5V, it will also take about 100mA at 5V. This also means that theoretically you can only power up to 100mA @ 1.5V. If you assume that the enumeration thing is not a problem, it would be 500mA @ 1.5V.
So, question 4: no, if you want that you need a switch mode power supply (SMPS). A linear supply (like a LM317) will have Current in = Current out, (leaving quiescent current out for now).
The switching power supply will try to be Power in = Power out (without it's efficiency taken into account).
So 5V 1A could be 2.5V 2A, 1.2 4.166A, as they all equal 5W. If you take into account an efficiency of 80%, you would probably see something more like 1.2 3A or something.
A SMPS is more complex to build up as it needs a inductor and a flyback diode. Also note that it may create a hum in the sound if the switching frequencies would 'leak' into your audio signal. So I think it's best to see if a LM317 is capable of powering your circuit.
@xsari3x: A current mirror is not used to deliver power. It's used to bias transistor amplifiers within opamps or other signal amplifiers. Furthermore, those outputs are constant current , where we need a constant voltage output here.
However I'd like a solution that can sustain 4.5V regardless of the
current (within the limits above)
If you are happy with a constant voltage of 4.5V then you should consider a low drop out voltage regulator like this: -
It can be setup to deliver up to 4.5V and at 1A to the load and will only need 4.645V (typically) on the input to sustain the output voltage. Note that it needs a minimum load for it work correctly (1mA) but this shouldn't be a problem given your requirements.
This is typical of a series of many voltage regulators with low drop-out voltage - I'm not saying use this one - I am saying be aware of what TI and other folk may be supplying. It's likely you can find one with current limit circuits built-in. You could also apply a bit of current limiting in this device by having a resistor in series with the voltage feed - this should be chosen to develop just enough voltage across it (at the required load current) to sustain the output. Should load current increase, output voltage will drop.
Best Answer
Consider Atmel's ATtiny85 with V-USB. It's an 8 pin AVR chip that you would have to program with V-USB, which is a software-level USB implementation that would enable the Enumeration Phase, which would allow you to use the entire 500mA available.
It's about a $1.50 in SOIC packaging, pictured below, which saves both space and cost:
Easily programmable and inexpensive, whereas the FTDI chip above (FT232R) is about $6 for one.
If you want to communicate with the chip, using V-USB also gives you the ability to act as a CDC-class USB device, which is akin to a serial port (UART), just like the FTDI chip.