There is no closed formula. Computing the number of graphs is painful. Probably it is best to do as bof does in the comments and look the answer up on OEIS.
Given a permutation $\sigma$ of $n$ vertices, let $X^\sigma$ denote the set of all labeled graphs that are unchanged when we apply $\sigma$. Then Burnside's lemma says that the number of unlabeled graphs is given by the calculation
$$
\frac{1}{n!} \sum_{\sigma \in S_n} |X^\sigma|.
$$
To compute $|X^\sigma|$, we reason from the cycle decomposition.
If $\sigma$ has a cycle of length $k$, the number of ways to choose the edges between vertices of that cycle is $2^{\lfloor k/2 \rfloor}$. (If $v_1, v_2, \dots, v_k$ are the vertices of the cycle, then once you decide if edge $v_1 v_i$ is present for $2 \le i \le \lfloor \frac k2\rfloor +1$, this determines the behavior of edges $v_2 v_{i+1}$, $v_3 v_{i+2}$, and so on.)
For every two cycles of lengths $k$ and $l$ in $\sigma$, the number of ways to choose the edges between their vertices is $2^{\gcd(k,l)}$ for similar reasons. If the first cycle is on vertices $v_1, \dots, v_k$ and the second on vertices $w_1, \dots, w_l$, then the behavior of edge $v_1 w_i$ is the same as that of edge $v_2 w_{i+1}$, $v_3 w_{i+2}$, and so on. This gives $\text{lcm}(k,l)$ different edges determined by $v_1 w_i$, and so there are $\frac{kl}{\text{lcm}(k,l)} = \gcd(k,l)$ binary choices to make.
So, for instance, the number of $9$-vertex graphs fixed by the permutation $(1\;2)\;(3\;9\;6)\;(4\;8\;7)\;(5)$ is
$$
2^{\lfloor 2/2\rfloor + \lfloor 3/2\rfloor + \lfloor 3/2\rfloor + \lfloor 1/2\rfloor + \gcd(2,3) + \gcd(2,3) + \gcd(2,1) + \gcd(3,3) + \gcd(3,1) + \gcd(3,1)} = 2^{11}.
$$
Because we only care about the lengths of the cycles, this is also the number of $9$-vertex graphs fixed by $(1\;2)\;(3\;4\;5)\;(6\;7\;8)\;(9)$ or any other permutation in the same conjugacy class. So instead of summing over all permutations, we can sum over all partitions $\lambda$ of $n$. We need to know the number $N(\lambda)$ of permutations in that conjugacy class; if $\lambda$ has $a_i$ parts of size $i$ (so that the permutations have $a_i$ cycles of length $i$) then there are $N(\lambda) = \frac{n!}{\prod_i (a_i)^i \cdot a_i!}$ such permutations.
Altogether, we get a sort of formula: the number of unlabeled graphs is
$$
\sum_{\lambda \mathbin{\vdash} n} \frac{N(\lambda)}{n!} \prod_i 2^{\lfloor \lambda_i/2 \rfloor} \prod_{i < j} 2^{\gcd(\lambda_i, \lambda_j)}.
$$
For example, when $n=4$, the partitions of $n$ are $(4)$, $(3,1)$, $(2,2)$, $(2,1,1)$, and $(1,1,1,1)$, and we have a term for each of them.
- $\lambda = (4)$ corresponds to $\frac{4!}{4} = 6$ permutations, and $|X^\sigma| = 2^{4/2} = 4$ for each of them.
- $\lambda = (3,1)$ corresponds to $\frac{4!}{3\cdot1} = 8$ permutations, and $|X^{\sigma}| = 2^{\lfloor 3/2\rfloor + \lfloor{1/2}\rfloor + \gcd(3,1)} = 4$ for each of them.
- $\lambda = (2,2)$ corresponds to $\frac{4!}{2^2\cdot 2!} = 3$ permutations, and $|X^\sigma| = 2^{\lfloor{2/2}\rfloor + \lfloor 2/2\rfloor + \gcd(2,2)} = 16$ for each of them.
- $\lambda = (2,1,1)$ corresponds to $\frac{4!}{2\cdot 1\cdot2!} = 6$ permutations, and $|X^\sigma| = 2^{\lfloor{2/2}\rfloor + 2\lfloor{1/2}\rfloor + 2\gcd(2,1) + \gcd(1,1)} = 16$ for each of them.
- $\lambda = (1,1,1,1)$ corresponds to only $1$ permutation (the identity) and $|X^{\sigma}| = 2^{6\gcd(1,1)} = 64$ for it.
Altogether we have $\frac{1}{4!}(6\cdot4 + 8\cdot 4 + 3\cdot 16 + 6\cdot 16 + 64) = 11$ graphs.
(Asymptotically, the last term, corresponding to the identity permutation, will be the dominant one for large $n$, and we get around $2^{\binom n2}/n!$ graphs. In other words, most graphs are completely asymmetric, and we can just pretend all of them are without too much error.)
