This is a follow-up to Graphs with many triangles but few complete graphs on 4 vertices

I'm looking for an upper bound for the difference between the number of edges and the number of 4-cliques in a graph on $n$ vertices subject to the condition that every edge is contained in a 4-clique. The best construction I know gives

$$\text{#edges} - \text{#4-cliques} = \frac{(3+o(1))n^2}{16}\approx 0.1875n^2$$

(take a complete bipartite graph with equal parts and add a matching in each part) and my guess is that this is best possible. But annoyingly, I can only prove

$$\text{#edges} - \text{#4-cliques} \leqslant \left(\frac{2(39+\sqrt{21})}{375}+o(1)\right)n^2\approx 0.232441n^2,$$

and even this weak bound requires a bit of work (and the removal lemma). Note that the $3/16$-result would follow (at least asymptotically) if one could prove that the number of triangles in an extremal graph is $o(n^3)$, which was the motivation of my previous question.

Has this type of problem been considered somewhere in the literature?

Do I miss something obvious, either in terms of a better construction or a better upper bound?

Are there reasons indicating that improving the bound is hard (besides me spending quite some time on trying)?