The exact power-of-seven bound for daisy-free triple systems

Let $\mathcal A$ be a family of $3$-sets of $[n]$, where $n=7^k$, and suppose that $\mathcal A$ contains no daisy. The power-of-seven daisy bound.

$$|\mathcal A|\leq \left(1-\frac{1}{49^k}\right)\frac{n^3}{12}=\frac12\binom{n+1}{3}.$$

This is proposed as the exact upper bound matching the iterated complement-of-the-Fano-plane construction, at least for orders that are powers of $7$; it would imply the 3-daisy density conjecture.