This quest has its impetus in a paper by Stanley and Zanello. I became curious about

*What is the sum of all hooks lengths of all partitions that fit
inside the $n$-th staircase partition?*

On the basis of experimental data, I'm prompted to ask:

Question.Let $\lambda=(n,n-1,\dots,1)$ be the staircase partition, $h_{\square}$ the hook-length of a cell $\square$ in the Young diagram of a partition. Then,

$$\sum_{\mu\subset\lambda}\sum_{\square\in\mu}h_{\square}=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}.$$ Is this true? If so, any proof?

**Update.** I'm still hoping for a complete solution for this evaluation; it's not common to get such a nice closed from in general. Part II of this problem has received a fine answer.

**Update.** To help Stanley's mention of a recurrence, denote the RHS quantity by $w_n:=\frac{(n+4)(2n+3)}6\binom{2n+2}{n+1} - (n+2) 2^{2n+1}$ then we have ("shaloshable")
$$n(n-2)w_n-2(4n^2-5n-3)w_{n-1}+8n(2n-1)w_{n-2}=0.$$