davivid:
I intend to drive all 16 channels at once, with 5A being more than required (I estimate most channels will need 2-4A). By inserting a 1k resistor in series with the base of each TIP107, will this not drastically lower the TIP107 output?
The TIP107 (or TIP105 or TIP106) has a minimum current gain of 1000 at 3A, and 200 at 8A. Therefore the minimim drive current needed for a 3A load is 3mA. I'm suggesting you use 10mA drive current, which should be enough for 5A with some margin. At 10mA the 1K resistors will drop 10v, but since you have around 12v available, they won't reduce the drive current - but they will dissipate most of the power that otherwise the 5940 would have to handle.
The other possibility I considered was to use P-channel MOSFETs instead of the TIP107s. You could connect each output pin of the 5940 to the gate of a IRF9540 mosfet and through a 1K resistor to +12v, connect the source of the mosfet to +12v, and the drain to the +ve side of the LED module. The advantage of using mosfets is that they drop much less voltage than darlingtons when on, so they run cooler and you only need 12v rather than 14v or so. The disadvantage is that with 1K driving resistance, they will take several microseconds to switch, which is OK if your PWM rate is not too high, but might otherwise cause them to get hot after all. You can avoid this using either a mosfet driver ic or a pair of small-signal transistors per mosfet, but this adds complexity.