How should I decide the cut-off frequency for my low-pass filter?
That looks like the spectrum of an 8-bit sound that is playing low bass notes only.  Most everything above that looks like noise floor.  (You can only get about 6 dB range per bit of resolution, so 8 bits gives about 48 dB max.)  If this is the case, you definitely don't want to reproduce the upper frequencies.  If you want to stick to your original PWM scheme, use pins 5 or 6 because they use 980 Hz PWM instead of 490.  I suggest you determine the filter by trial and error, since it comes down to how much PWM carrier you can tolerate.  You can make a simple RC filter using the formula F = 1 / (2 * pi * R * C) where F is the cutoff frequency in Hz, R is ohms, and C is farads (that's 10^6 microfarads per farad).  This type of simple filter will have a very soft cutoff frequency, about 6 dB per octave, so if you set it for 123 Hz (3 octaves below 980 Hz) the 980 Hz PWM will be reduced by 18 dB.  The R connects to the PWM pin, and the C goes from the other end of R to ground, as does the signal output.  The formula assumes that there is no output load, so you should connect an amplifier there.  If you are like most folks, you probably have fewer capacitor values on hand than resistor values, so pick the C value first.  Let's say it's 1 uFd.  Then if you use R=1K the cutoff will be 159 Hz.  Change R while listening to the output