C1 = 1 / (2 * pi * 5.00025e6 * 20,000)
1.59pF! They don't even make caps that small do they?!
This equation would be correct - if you wanted to exclude all frequencies below 20khz.
Actually, I think you want to exclude all frequencies below maybe 40Hz or 75Hz and pass the rest through.
So if you wanted to pass bass guitar at 40Hz, you might calculate
C1 = 1/ (2*pi*5E6*40) = 796E-12 = 796pF; might as well use 1000pf.
If you want to exclude everything over 20kHz, you do that with C2. In this case, the 1k resistor predominates and
C2 = 1/(2*pi*1E3*20E3) = 7nF = 7000pF.
An interesting point on interactions. With the C1/R1/R3 stuff mixed up with the R3/C3 filter, this will not work as expected because the C1/1 network is loaded down by R2/C2. The impedance level of R2/C2 is too low.
If you rearrange the network so the input signal hits R2/1K first, is shunted to ground by C2 for high pass removal, then connects to R1 to ground and C1 in series to R3, then it works as intended. R2/C2 removes the highs, and is a low impedance source for the very high impedance C1/R1/R3 high pass section.