There are $\binom{n}2=\frac12n(n-1)$ pairs of distinct points. If you do not allow loops or multiple edges, each of these pairs determines one possible edge, and you can have any subset of those possible edges. A set with $\binom{n}2$ members has $2^{\binom{n}2}$ subsets, so there are $2^{\binom{n}2}$ possible graphs without loops or multiple edges.
If you demand that the graphs be connected, the problem becomes very much harder. From your final comment I take it that you are in effect counting labelled graphs. This sequence of numbers is A001187 in the On-Line Encyclopedia of Integer Sequences. If $d_n$ is the number of labelled, connected, simple graphs on $n$ vertices, the numbers $d_n$ satisfy the recurrence
$$\sum_k\binom{n}kkd_k2^{\binom{n-k}2}=n2^\binom{n}2\;,$$
from which it’s possible to calculate $d_n$ for small values of $n$. This recurrence is derived as formula (3.10.2) in Herbert S. Wilf, generatingfunctionology, 2nd edition, which is available for free download here.
According to MathWorld, Brendan McKay’s software package nauty includes a routine that efficiently enumerates such graphs; it’s available here.
If you count unlabelled graphs instead, so that you don’t count isomorphic graphs separately, you get the sequence mentioned by Arturo in the comments.
The sum of degrees is $10$, and since the graph is self complementary, by symmetry the degree sequence must be
$$(d_1,d_2, 2,4-d_2,4-d_1) \,,$$
where $d_1, d_2 \in \{ 0,1,2 \}$, and $d_1 \leq d_2$.
It is easy to see that $d_1=0$ is not possible, since then the last degree would be $4$, thus $d_1, d_2 \in \{ 1,2 \}$, and $d_1 \leq d_2$.
Case $1$ $d_1=1, d_2=1$. Your degree sequence is $(1,1,2,3,3)$. Each of the vertices of degree $3$ must be connected to all vertices excluding one end vertex. There is only one graph up to isomorphism.
Case $2$ $d_1=1, d_2=2$. Your degree sequence is $(1,2,2,2,3)$. You have two graphs here: the end vertex is connected to a degree $2$ or $3$.
Case $3$ $d_1=2, d_2=2$. Your graph is connected (why?) and has all degrees $2$. Only 1 possibility.
Best Answer
There are $2^{\binom{n}{2}}$ labeled graphs on the vertex set $\{1,2,\ldots,n\}$ and each decomposes into:
a $k$-vertex connected component $K$ containing vertex $1$, for some $k \in \{1,2,\ldots,n\}$, and
one of the $2^{\binom{n-k}{2}}$ subgraphs on the vertices disjoint from $K$.
Hence, if $C_n$ denotes the number of $n$-vertex labeled connected graphs, then $$2^{\binom{n}{2}}=\sum_{k=1}^n \binom{n-1}{k-1} C_k\, 2^{\binom{n-k}{2}}$$ and $C_1=1$.
In small cases, we have \begin{align*} 2 &= C_1 + C_2 \\ 8 &= 2 C_1 + 2 C_2 + C_3 \\ 64 &= 8 C_1 + 6 C_2 + 3 C_3 + C_4 \\ 1024 &= 64 C_1 + 32 C_2 + 12 C_3 + 4 C_4 + C_5 \\ \end{align*} which can be used to determine $C_2,\ldots,C_5$.
By the way, while it might be theoretically possible to count them by drawing them by hand, it's going to be tedious. For $4$-vertex graphs, below lists the non-isomorphic connected graphs along with the number of labeled graphs isomorphic to it:
This gives the $38$ connected labeled graphs on $4$ vertices; there's apparently $728$ on $5$ vertices (see Sloane's A001187).