The skin depth equation may not be reliable at such low frequencies, but let's pretend that it is. Skin depth is about 0.82" at 10 Hz and 0.018" at 20 kHz. The area of a circle is A = pi*r^2 and for an anulus it is A = pi*(R^2 - r^2) where R is the outer radius and r the inner radius.
12 AWG wire is about 0.081" in diameter and the DC resistance (DCR) is about 1.6 m-ohm/foot, or about 0.016 ohms for 10', and thus about 0.032 ohms for a speaker cable (two wires). The capacitance of a pair depends upon the insulation thickness and construction style (some spae the conductors further apart, reducing the capacitance). A quick look shows a range of about 10 to 20 pF/foot, compared to about 30 pF/foot for a 75-ohm coax interconnect, so that range seems reasonable (about half that of coax). I happen to have a reference showing 19.5 pF/foot for shielded 12 AWG wire so let's use 15 pF for typical unshielded 12 AWG speaker cables (middle of the range). That is 150 pF for a 10' pair.
At 10 Hz the skin depth exceeds the cable's diameter so the entire cable is used, and DCR = ACR = 32 m-ohms. The area is A = pi*(0.081/2)^2 = 0.00515 sq inch for a single wire. We can ratio that to the anulus area to estimate the resistance of the anulus.
At 20 kHz I'll use the skin depth to find the cross-section area A = pi*[(0.081/2)^2 - (0.081/2-0.018)^2] = 0.00356 sq inch. The outer radius is the entire wire as above, and the inner radius is the outer radius less the skin depth.
Now the pair of 12 AWG wires 10' long is 32 m-ohms (0.032 ohms) at DC and the same at 10 Hz. At 20 kHz, we can ratio the cross-sectional area to estimate the resistance at 20 kHz as 0.032 ohms * 0.00515/0.00356 = 0.0463 ohms. For marketing, that is a 45% increase in resistance, or 3.2 dB. Seems huge, eh?
But look at the real-world implications... Aside from the output impedance of the amp, which is 80 m-ohms for a damping factor of 100 (and usually much, much lower at 20 kHz so output impedance is much, much higher), there is the crossover and speaker impedance in the picture, all of which are likely much higher than that of the wire at either frequency. But since frequency response is commonly cited, we can look at that.
I'll do a lumped model because it is easier for a back-of-the-envelope calculation. In the real world, resistance and capacitance is distributed along the length of cable, so my simple model will be slightly off. Feel free to fire up your EM simulator or measure on a VNA if you want a closer result.
Conductance is at the surface so capacitance to first order is the same at either frequency. At 10 Hz, the effective cable time constant is R*C so 0.032 ohms * 150 pF = 4.8 ps. At 20 kHz, it has risen to 0.0463 ohms * 150 pF = 6.9 ps, an increase of 2.1 ps. A ps is 10^-12 seconds. If your system has 100 kHz bandwidth, that equates to a first-order time constant of around 2.2/(2*pi*f) = 2.2/(2*pi*100e3) = 3.5 us, about 435,783 times larger than the change due to skin effect in the cable. I am pretty sure I could not hear the difference. I started to calculate phase shift and such but it's just not worth my time and I have to get back to work.
Check my math, I was in a hurry - Don