Electronic – Can someone put a name to this DC-coupled low pass filter topology

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.

If you use this circuit on a regular basis I guess it would be worth to calculate its transfer function once and for all, so you can just easily evaluate it given the concrete component values. Doing the math you get (assuming an ideal OP)

$$H(\omega)=\frac{j\omega R_1C_1}{1+j\omega R_1C_1}\left(1+\frac{j\omega R_4C_2}{1+j\omega R_2C_2}\right)\tag{1}$$

A plot of (1) in dB with the specified component values looks like this (|H|/db vs f in Hz):

from which you can see that the -3dB point is at about \$250\,\text{Hz}\$.

If you further assume that

$$R_1C_1=R_2C_2=\tau$$

and if you denote the gain at large frequencies by

$$g=1+\frac{R_4}{R_2}$$

then with a bit more math you get this exact expression for the 3dB cut-off frequency in radians

$$\omega_c=\frac{1}{\tau}\sqrt{1-\frac{1}{g^2}+\sqrt{\left(1-\frac{1}{g^2}\right)^2+1}}\tag{2}$$

which with the given component values gives

$$\omega_c=2\pi\cdot 247.28$$

For a large gain \$g\gg 1\$, formula (2) is closely approximated by

$$\omega_c=\frac{1}{\tau}\sqrt{1+\sqrt{2}}\approx\frac{1.55}{\tau}\tag{3}$$