When you have a synchronous buck converter, the input to the LC stage is basically a "hard" square wave and this is averaged by the LC like so: -
Then, to reasonably calculate the output ripple voltage, consider the LC as a low pass filter and set the natural resonant frequency to be low enough to attenuate the 50 kHz square wave by a decent amount. If you chose the resonant frequency to be (say) 500 Hz there would be approximately 80 dB of attenuation at 50 kHz: -
Interactive filter calculator.
I've chosen values for L and C and R that produce a resonant frequency of about 500 Hz and R would be about 0.5 ohms for the required power throughput quoted. If you use the controls you can find that the attenuation at 50 kHz is 80 dB.
That 80 db is the reduction in the AC signal content that is produced by the switching action. So, if the AC content was a 20 volts peak-to-peak square wave, this could be reasonably approximated by a sinewave of about 25 volts peak-to-peak. This is attenuated by 80 dB (10,000 times) to produce an output ripple voltage of 2.5 mV peak-to-peak.
So, decide what ripple voltage is acceptable to meet your design aims and choose the output capacitor to work with the inductor to give you enough attenuation.