Electronic – Cut-off frequencies for common mode and differential mode filter

The -3dB point is your cutoff frequency. It's just standard practice to define it that way. In order to find what your values should be, I'd go with equal element implementation (it's simpler, and you can correct for gain with a simple gain stage later if you need to). Choose R1=R2, C1=C2, and pick a value for either R or C. This yields the following formula for the cutoff frequency: $$f_0=\frac{1}{2\pi RC}$$ I generally choose a value of C initially, as it's easier to find or make a resistor with a strange value, whereas it's more difficult with capacitors. So, set your cutoff frequency equal to f0, and solve for R.

Here's an example: let's say I want a LPF with f0=250 Hz. I'll choose C to be 0.1 micro and solve for R. $$250=\frac{1}{2\pi RC} \rightarrow 250=\frac{1}{2\pi R(0.1x10^{-6})}\rightarrow R\approx6400\Omega.$$

From there, all you need to do is implement your circuit. Once you know what your value for R is supposed to be, you can use a dual-channel potentiometer that has the correct resistance within it's range in place of the two resistors (for the above example, something like a 10k ohm potentiometer would do the trick). This will allow you to change your cutoff frequency, since it's based upon both R and C.

Edit: As Matt Young suggested in the comments, adding a resistor in series with the potentiometer will set the maximum cutoff, and prevent shorts. It's an excellent addition to the circuit, and will keep some sanity when adding the potentiometers.

The natural resonant frequency of a 2nd order low pass filter will have a pole zero diagram like this: -

If you are unfamiliar with pole-zero diagrams see if this helps: -

If you follow that, then go back to the first diagram and you should realize that the undamped natural resonant frequency (\$\omega_0\$) has a value anywhere on the semi circle and it is the Q of the filter how far round from the jw axis the two poles are. Given that any order of butterworth filter has all its poles on the same semi circle, the answer to your question is: -

Is the cut-off frequency for each high order butterworth filter the same or different?

If you mean the natural resonant frequency then YES!

If you mean the 3dB point of each filter's response on the jw axis (the axis that pertains to "real-life" measurements on a spectrum analyser) then NO!

Butterworth pole-zero diagram: -

The above shows the poles of a 10th order butterworth filter - note that all the poles lie on the same circle and therefore all the individual filters (5 x 2nd order) have the same natural resonant frequency.

Taken from here.