Is there a closed form formula for counting 2-regular labelled graphs? Do we have a closed form formula for counting undirected 2-regular labelled graphs ?
The sequence for there enumeration is given here.
 A: Initially the question said "$2$-regular graphs".  It was changed to labelled graphs after I had written this answer.  So, the first part of the answer addresses unlabelled graphs, and the second part (after the EDIT) explains why I think it would be hard to extend this to labelled graphs.
A finite $2$-regular graph is a disjoint union of cycles, so the number of $2$-regular graphs on n vertices is the number of partitions of n into parts of size$\geq 3$.
As far as I can determine, there isn't a closed form known for this.  At least, I haven't found one by searching the Web.  One can give a generating function of course.  If $Q(n)$ is the number of partitions of $n$ into parts of size $\geq3$ then $Q(n)$ is the coefficient of $x^n$ in $$\prod_{k=3}^\infty\frac1{1-x^k}=(1-x)(1-x^2)\prod_{k=1}^\infty\frac1{1-x^k}$$
The last infinite product is the generating function for the ordinary partition function, so if $p(n)$ is the number of partitions of $n$, then $$Q(n)=p(n)-p(n-1)-p(n-2)+p(n-3)$$
We can also see this by inclusion and exclusion.  To get the partitions into parts $\geq3$ we must exclude the partitions with a $1$ and the partitions with a $2$, but the partitions with both a $1$ and a $2$ must be added back in.
A recursive formula for the partition formula is known of course.  (See the last section these notes.)
EDIT
Offhand, the case of labelled graphs sounds harder.  Take the $n=8$ case.  There are three $2$-regular graphs:
$$C_8\\
C_4\cup C_4\\
C_5\cup C_3$$
where $C_k$ is a $k$-cycle.  There are $\frac{7!}2$ labelled graphs corresponding to the first case.  For the second case, there $\binom84$ ways to choose the labels of the first cycle, and then $(3)^2$ ways to label the vertices, but we must divide by $2$, to take account of isomorphism of the two cycles, so $\frac 12\binom84(3)^2$ labellings.  In the third case, we have $\binom83\cdot12$ labellings.
This approach would require listing all the partitions of parts of size $\geq3$.
