Switching power supplies with a relay, how to account for delay between connections?
A capacitor is probably not the best way to approach this problem. A solution using FETs probably is. However, here's how you'd size the capacitor if you actually did want to go the capacitor route.

The fundamental property of a capacitor that relates current and voltage to capacitance is:
$$I=C\frac{dV}{dt}$$
In other words, one amp of current flowing through a one Farad capacitor will change the voltage across that capacitor by one volt per second. Any amount of current flow will result in a change in voltage. Therefore, you need to decide how much voltage change you can live with. You may need to experiment to see how far the voltage on the power supply can drop before the PC reacts badly (browns out, powers off, locks up, etc).

Let's assume that the PC can handle a temporary voltage droop down to 8V. In the worst case, the PC is consuming a constant 20A from the supply. And the worst case relay deadtime is 50ms. Plugging into the equation above:
$$20A = C\frac{(12V-8V)}{50ms}$$
Solve for C:
$$C=0.25F$$
That translates to: A 0.25F capacitor charged to 12V that is suddenly cut off from the power source can supply 20A to your PC for 50ms and drop in voltage by 4V during that time.

The math part is easy. Sourcing an actual capacitor that big (with an appropriate voltage rating) that won't blow your budget may be hard.