If $cd(S_n)$ is the set of degrees of irreducible, complex-valued characters on the symmetric group $S_n$, then by the basic representation theory of the symmetric group this is the same as the set $\{f^{\lambda}\colon \lambda\vdash n\}$, where for a partition $\lambda\vdash n$, $f^{\lambda}$ is the number of Standard Young Tableaux of shape $\lambda$, which by the hook-length formula has the explicit form $f^{\lambda}=n! \cdot \prod_{u \in \lambda}\frac{1}{h_u!}$. I would imagine that the number of pairs of partitions $\lambda,\nu \vdash n$ with $f^{\lambda}=f^{\nu}$ is very small as a fraction of all pairs of partitions of $n$. And probably that would not be hard to formally establish (for instance using the hook-length formula). So the size $|cd(S_n)|$ of the set of degrees of irreps should be basically the same as the number of partitions on $n$. This is a very famous quantity whose asymptotics are well-understood (in particular the partition number is known grow super-polynomially; for a starting point see e.g. the Wikipedia entry https://en.wikipedia.org/wiki/Partition_function_(number_theory)).

EDIT: As pointed out in the comments below, there are some collisions you do have to worry about if you want to understand precise asymptotics (e.g. a partition and its transpose have the same number of SYT), but nevertheless if you just want to establish that $|cd(S_n)|$ grows super-polynomially, surely this naive reasoning should be enough.

EDIT 2: In the comments a heuristic why there should be few collisions was requested. A simple heuristic why there should be few collisions of values of $f^\lambda$ is that the maximum value of $f^\lambda$ over $\lambda \vdash n$ is on the order of $\sqrt{n!}e^{-\alpha\sqrt{n}}$ (see e.g. https://arxiv.org/abs/1804.04693), while $f^\lambda$ can be as small as one, so the range of values taken on by $f^\lambda$ is quite large (compared to say the number of partitions of $n$).