Based on this data sheet for the 74HCT173, which section(s) describes the maximum current I can draw per output pin?
There are some considerations here.

Maximum ratings.

The absolute maximum ratings must not be exceeded (and you should not normally go anywhere near them).

For this device, they are:


This shows that you could draw up to 35 mA from an output, but the total device current must not exceed 70 mA (so you could not draw this from all the outputs). In addition, you need to keep the device reasonably cool so you would want to keep the power dissipation down.

Now we can look at the output current vs. output voltage:



If I assume you are using the device with a 5 V supply and driving the outputs low to run your LEDs (not listed here so I will use the 4.5 V numbers), then if you draw 6 mA, at room temperature you will have an output voltage of between 16 mA and 26 mV.

That yields power in the output stage of 156 &micro;W per output so the total would be 624 &micro;W if all 4 were driven low at the same time.

Now 6 mA is usually quite sufficient for most LEDs in normal lighting.

If you want to drive a bit harder, then although the data sheet is silent on what the output voltage would be at 10 mA, we could estimate it. The output resistance of the bottom transistor is about $$\frac {26\ mV} {6\ mA} = 4.3\ \Omega.$$.

At an output current of 10 mA, there would be an output voltage of 43 mV for a power dissipation of 430 &micro;W per output.

So it might be possible to drive 15 mA (total 60 mA if all outputs are active) which would yield about 640 &micro;W per output; that would give a total current of 60 mA if all outputs were active which is within (but a bit close for my taste) the absolute maximum ratings. You will need to keep in mind that there are other currents that add to Icc (see 'additional supply current' on page 7).

Note that if the outputs are switching at any sort of fast rate you will need to add the power dissipated by switching (the calculation is on page 10 of the datasheet).