In a stack of N dominos, the following recursive requirement must be met:

For any M dominos at the top of the stack, their total center-of-gravity must not be past the edge of the domino/table directly beneath it.

So let N be infinity. Pretend the stack is straight (there is no overhang). Now, let's start pushing out the dominos starting at the top.

(M=1) The topmost domino, since its center of gravity is at half its length, can be pushed past the edge of the domino below it by half its length.

(M=2) Since the first domino extends past the edge of the second domino by 1/2, the top two dominos has its center-of-gravity at 1/2 + (1/2)(1/2) = 3/4 the length of the second domino. Thus, the second domino can be pushed out by 1/4 of its length.

(M=3) Since the center-of-gravity of the first two dominos extends past the edge of the third domino by 1/2, the center-of-gravity of the top three dominos is at 1/2 + (2/3)(1/2) = 5/6 the length of the third domino, so the third domino can be pushed out by 1/6 of its length.

(M=4) Since the center-of-gravity of the first three dominos extends past the edge of the fourth domino by 1/2, the center-of-gravity of the top four dominos is at 1/2 + (3/4)(1/2) = 7/8 the length of the fourth domino, so the fourth domino can be pushed out by 1/8 of its length.

...

...

(For any M) Since the center-of-gravity of the first M-1 dominos extends past the edge of the Mth domino by 1/2, the center-of-gravity of the top M dominos is at 1/2 + ((M-1)/M)(1/2) = 1-(1/2M) the length of the Mth domino, so the Mth domino can be pushed out by 1/2M of its length.

Thus, the total overhang of an infinite stack of dominos is half the sum of the harmonic series (1+1/2+1/3+1/4+1/5+...), which is infinite.

Of course, you wouldn't actually be able to make a infinite overhang with "real" dominos.