Three-run enumeration conjecture for flattened Stirling permutations

Let $\mathcal{Q}_n$ denote the set of Stirling permutations of order $n$, and let $\operatorname{flat}_3(\mathcal{Q}_n)$ be the set of those with exactly three runs. Three-run enumeration conjecture. The number of Stirling permutations of order $n$ with exactly three runs is

$$|\operatorname{flat}_3(\mathcal{Q}_n)| = 2\sum\limits_{k = 1}^{n-1} \binom{n-1}{k} \left( \sum\limits_{j = 2}^{n-1 - k} \binom{n-1 -k}{j} \right) + \sum\limits_{k = 2}^{n-1} \binom{n-1}{k} \left( \sum\limits_{j = 2}^{n-1 - k} \binom{n-1 -k}{j} \right) + \sum\limits_{k = 3}^{n-1} (2^{k-1} - 2)\binom{n-1}{k}.$$

This was computationally verified for $1\leq n\leq 12$, but a proof is not supplied.