This circuit will pull 350ma for just the LED, in addition to more from the series resistor.
Um, no: if there's 350mA going through the LED, that same current is flowing through the resistor.
But there probably isn't: the TIP120 is going to be losing about 1.2-1.4V because it's a Darlington. As I recall, those 1W Luxeons have a nominal forward voltage of about 3V (but do not bet on my vague recollection being correct). That means a drop of about .6-.8V across the resistor, so you'd only be getting about 120-150mA.
So you're probably being overly conservative in your resistor choice, and may be able to get brighter flashes if you want.
If you do want brighter flashes, go back to the datasheets. I recommend assuming the worst cases on the TIP120 (1.2V) and the LED (whatever the minimum forward voltage is), and actually measuring the 5V supply. Then you can probably reduce the resistor value to get a voltage drop that gives you the full 350mA.
You might even be able to go over 350mA, depending on what the spec sheet says, if you keep the flash length short enough and the duty cycle low. But then you have to bet that you never have a software glitch that leaves the LED on long enough to fry. You have to ask yourself: Do you feel lucky, Punkin? 