One known fact is that if $\sigma\in S_n$ is an $m$-cycle, then $|C_{S_n}(\sigma)| = m\cdot(n-m)!$.
From this equation, we can compute the Class equation easily.
My question is that can we obtain some general formula for arbitrary $\sigma\in S_n$?
For example, $|C_{S_n}(\sigma)|$ where $\sigma$ is a product of disjoint $m_1,m_2$ cycles (cycle decomposition).
Recall the fact that if $\sigma\in S_n$ has the cycle decomposition $$(a_1\ \cdots\ a_{k_1})(b_1\ \cdots\ b_{k_2})\cdots,$$ and $\tau\in S_n$, then $$\tau\sigma\tau^{-1} = (\tau(a_1)\cdots\tau(a_{k_1}))(\tau(b_1)\cdots\tau(b_{k_2}))\cdots.$$ Since $C_{S_n}(\sigma) = \{\tau\in S_n\mid\tau\sigma\tau^{-1} =\sigma\}$, you can find the order of $|C_{S_n}(\sigma)|$ easier than you think, in practice.
For example, to compute the order of $C_{S_n}((1\ 2)(3\ 4))$, you need to find the bijection $\tau\in S_n$ such that $(\tau(1)\ \tau(2))(\tau(3)\ \tau(4)) = (1\ 2)(3\ 4)$. Noting the above fact, can you see that $|C_{S_n}((1\ 2)(3\ 4))| = 8\cdot(n-4)!$ for $n\geq 4$?
If you want to compute the centralizer of $\sigma\in S_n$ to compute the Class equation, actually there is a formula for $S_n$ that counts the number of conjugates of $\sigma$.
Let $\sigma\in S_n$ and $m_1,...,m_s$ be the distinct integers that appear in the cycle type of $\sigma$ (including $1$-cycles) and for each $i\in \{1,...,n\}$, assume $\sigma$ has $k_i$ cycles of length $m_i$. Then the number of conjugates of $\sigma$ is $$\frac{n!}{(k_1!m_1^{k_1})\cdots(k_s!m_s^{k_s})}.$$