Depends on how you set it up. With your suggested series connection of the peotentiometer, the value is somewhat more critical and 50k might not be suitable.
The required resitance would depend on the input impedance (resistance) of the amplifier/headunit aux in/whatever.
If the potentiometer was of the same size as the input impedance, you can at max drop 50% of the voltage across the potentiometer - which means a 3dB attenuation. If you want more, then the potentiometer needs to be larger than the input impedance...

But the use suggested above isn't the "usual way". Instead hook up the signal source to the "ends" of the potentiometer and wire the amplifier to one end and the slider.

If you don't want it capable of turning it down to 0, you can attach a fixed resistance at one end - your max attenuation is now R/(P+R) times (recalc to dB not done...)
Something like this:
Code:

---

*-------
        |
        P
        P
        P<---------*
        P
        |           To input on amp
        R
        R
        |
*--------------------*

---

As to the power requirements... Say you've got a signal source that provides a 4V RMS signal. Power developed (P=U^2/R) into 50k is 4^2/50k = 16/50.000 = 0.00032W

For a true and proper power calculation one would have to consider the effects of the parallell connection with the amp input and whether parts of the potentiometer could handle things in whatever position the slider is in. But with 50k, the margins should be sufficient so that more detailed calculations can be avoided.