Electrical – How long will an 6000 mAh battery pack power a 12v 60W amp for

A 3.5 Volt, 1000 mAh battery can supply 3.5 watt hours. That means the battery can supply 500 watts for 0.007 hours. So, with perfect efficiency of all circuitry, including stepping up the battery's 3.5 volts to the 12-24V the motor needs, you could theoretically supply full power to the motor for 25 seconds.

Of course, it's very unlikely the battery could tolerate being drained in 25 seconds. The fastest draining batteries I've seen can tolerate being fully drained in 72 seconds with only moderate loss of lifespan.

You're going to have to design a circuit to step up the 3.5 volts from the battery to at least the 12V that seems to be the motor's minimum specification.

An 11.1V 30-60C LiPO battery (such as those used for drones) seems like a reasonable match for your application, and could probably run the motor at 1/2 voltage (just under 12V) and 1/2 duty (pulsed). The one I linked to stores about 24 watt hours and could probably run the motor at 1/3 speed or so for about 5 minutes. It's designed to tolerate complete discharge from a fully charged state in as little as one minute.

Update: Since this is a quadcopter application, you should be looking at using a brushless motor with an electronic speed controller. Otherwise, it will be very difficult to adjust the rotation rate of the motor quickly enough to provide a stable platform. You should look at the design of a quadcopter that's similar in size and weight to your requirements and look at what batteries, motors, speed controllers, and props it uses.

If a power supply says the input power is 100-240V, ~1.5A, I'm assuming that the "1.5A" draw would be at 100V, not at 240V, right? I.e. would the total max watts be 150W regardless of the input voltage?

Yes, the PSU input power performance will be given by

$$ P_{OUT} = P_{IN} \times efficiency $$

... the output rating on this supply is 19V, 3.42A.

$$ P_{OUT} = VI = 19 \times 3.42 \approx 70~W $$

If the efficiency is 85%, for example, the input power would be $$ P_{IN} = \frac {P_{OUT}}{efficiency} = \frac {70}{0.85} = 82~W $$

At 100 V current would be < 1 A.

The 1.5 A is worst case surge current. If the efficiency was only 50%, for example you would have 70 W dissipated in the power supply and it would be very hot to the touch. The main current should be much less than 1.5 A in normal operation.