It seems like, even if you are only interested in that second answer, you're going to running up against the problem of computing the number of connected graphs on $a_i$ vertices. There might be a better way to count them that this, but here's one possible formula:
Let $P_c(n)$ be the set of partitions of $n$ with $c$ parts. For a partition $P=a_1+a_2+\cdots+a_c$, define the following numbers. Let $m_r(P)$ be the number of parts of $P$ of size $r$, i.e.
$$m_r(P)=\\#\{a_i : a_i=r\}.$$
Let $\omega(P)=\prod_{r=1}^{n}\bigl(m_r(P)!\bigr)$. Finally, let ${\mathcal C}(a_i)$ be the number of connected graphs on $a_i$ vertices.
One formula for the number of graphs on $n$ vertices with $c$ connected components then is
$$\sum_{P\in P_c(n), P=a_1+a_2+\cdots a_c} \binom{n}{a_1,a_2,...,a_c}\frac{1}{\omega(P)}\cdot\prod_{i=1}^c {\mathcal C}(a_i).$$
The multichoose coefficient $\binom{n}{a_1,a_2,...,a_c}$ comes from the number of partitions of the vertices into $c$ groups, with $\omega(P)$ correcting for the overcount factor for multiple groups of vertices of the same size. The product comes from, for a fixed partition of vertices into pieces of size $a_i$, how many connected graphs are possible on each piece.
I don't believe there are explicit formulas for these $C(a_i)$, and I'm not sure that there's a way to generalize this to answer your original question (outside of in a painful fashion summing across ways divvying up the $e$ edges across these $c$ components.)
While this doesn't say that your question is as hard as computing the Tutte polynomial itself, it makes me suspect that there isn't a closed formula. I looked for a bit as well at Stirling numbers of the second kind (which counted the number of set partitions of the vertices into $c$ pieces), but didn't see a nice way to pull those apart to count the number of connected graphs possible given a set partition. That might give you a different formula though with an approach like that.
Best Answer
Something to start with.
We have $$\Theta(G)=\sum_{\alpha} (-1)^{\alpha}\#G_\alpha=\sum_{\alpha}(-1)^\alpha\sum_U \mathbf{1}(U\,\text{is a component of}\,G_{\alpha}),$$ where $U$ runs over all induced connected subgraphs of $G$. Changing the order of summation, we get $$ \Theta(G)=\sum_U \sum_{\alpha:\, U\,\text{is a component of}\,G_{\alpha}} (-1)^{\alpha}. $$ Look at the inner sum. Let $\tilde{U}$ denote the set of vertices $v\in V\setminus U$ (here $V=$set of vertices of $G$), which have a neighbour in $U$, and $U^*$ denote $V\setminus (U\cup \tilde{U})$. Then $G_\alpha$ must contain $U$, should not have vertices in $\tilde{U}$ and may contain arbitrary subset of $U^*$. It yields that the inner sum has $2^{|U^*|}$ summands, and that it vanishes unless $U^*=\emptyset$. In the latter case, the inner sum equals $(-1)^{|U|}$. Such $U$ for which $U^*=\emptyset$ (that is, every vertex either belongs to $U$ or has a neighbour in $U$) are called dominating. Thus $$ \Theta(G)=\sum_{U\,\text{is dominating and connected}} (-1)^{|U|}. $$ If $G=C_n$ is a cycle of length $n$, then $U$ should be either the whole cycle (1 variant), or a path of length $n-1$ ($n$ variants) or a path of length $n-2$ ($n$ variants), so we get $\Theta(C_n)=(-1)^n$.