This question is a continuation of a question asked yesterday which had a very nice answer.
Consider the summation $$\sum_{\lambda \subset (k)^n} \dim S_{\lambda^t} (\mathbb{C}^k) \cdot \dim S_{[\lambda]} (\mathbb{C}^{2n}) $$ where $\mathbb{C}^{2n}$ is a symplectic space and $S_{[\lambda]}$ instead denotes the symplectic Schur functor. Then, experimentally there is an equality $$\sum_{\lambda \subset (k)^n} \dim S_{\lambda^t} (\mathbb{C}^k) \cdot \dim S_{[\lambda]} (\mathbb{C}^{2n}) = 2^{n(k-1)} \cdot \sum_{\mu \subset (k+1)^n} \dim S_\mu (\mathbb{C}^n) $$ and in view of Sam Hopkins's answer to the linked question we thus have a closed form $$\sum_{\lambda \subset (k)^n} \dim S_{\lambda^t} (\mathbb{C}^k) \cdot \dim S_{[\lambda]} (\mathbb{C}^{2n}) = 2^{n(k-1)} \prod_{1 \leq i \leq j \leq n} \frac{k+i+j}{i+j-1}.$$ I am not quite sure how to prove this identity. The sum $\sum_{\mu \subset (k+1)^n} \dim S_\mu (\mathbb{C}^n)$ is equal to the dimension of the odd orthogonal Schur functor $S_{[(k+1)^n]}^B (\mathbb{C}^{2n+1})$ ($B$ for type $B$), which even at the level of characters agrees with the orthosymplectic Schur functor $S_{[(k+1)^n]} (\mathbb{C}^{2n|1})$. This is the only connection I can see with the fact that there are symplectic Schur functors appearing on lefthand side of the equality, but the additional power of $2$ is mysterious to me. It's possible that what is really happening here is that there's a property like $2^n \dim S_{[(k+1)^n]} (\mathbb{C}^{2n|1}) = \dim S_{[(k+2)^n]} (\mathbb{C}^{2n|2})$ (this type of property does hold for ranks of regular super-Schur functors), and this can be iterated to obtain something along the lines of an identity $$\dim S_{[(2k)^n]} (\mathbb{C}^{2n|k}) = \sum_{\lambda \subset (k)^n} \dim S_{\lambda^t} (\mathbb{C}^k) \cdot \dim S_{[\lambda]} (\mathbb{C}^{2n}),$$ which is admittedly an elegant/plausible identity (if true), but I am unable to prove it. Alternatively, there might just be a direct way to obtain the formula I mentioned above.
