Most pots do have DC through them. Its no different than any other resistor. Lets say its a 1/20 watt 5k pot and a 9v battery running that amp circuit. Also assume for a conservative estimate that Q2 is fully on and the 220ohm and 330ohm are the only other things in series with the pot for a total of 5550 ohm. 9v/5550=0.00162A. So we know the pot will never see more than 1.62mA of current through it. From this we also know the voltage across the 5k pot, which is 1.62mA*5k=8.11v. Now we know the voltage across it and the current through it, and therefore we know the power dissipated in it. 8.11v*1.62mA=13.13mW. 1/20w is 50mW, more than enough. The pot survives.
In real life, Q2 is not fully on, probably biased somewhere in the middle, and is adding resistance to that path further decreasing the current through it. That pot is nowhere near being damaged.
edit: The pot doesn't care if it is AC or DC. You give it enough power, it goes up in smoke. Ohms law applies exactly the same.