I would like to contribute some ideas even though I don't have as much
time as I'd like at the moment. If I understand this problem correctly
then the class of graphs under consideration call it $\mathcal{Q}$ is
in a set-of relationship with the class of endofunctions call it
$\mathcal{E}$ with the latter being sets of the former.
The following MSE link A has closely related material, as does this MSE link B where aspects of "Random Mapping Statistics" by Flajolet and Odlyzko are discussed.
This gives the species equation
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{E} = \textsc{SET}(\mathcal{Q})$$
which is in terms of generating functions
$$E(z) = \exp Q(z)
\quad\text{or}\quad
Q(z) = \log E(z).$$
Recall the popular labeled rooted tree function which represents the
species
$$\mathcal{T} = \mathcal{Z} \times \textsc{SET}(\mathcal{T})$$
and has the functional equation
$$T(z) = z \exp T(z).$$
We also have that $$T(z) = \sum_{n\ge 1} n^{n-1} \frac{z^n}{n!}$$
(Cayley's formula)
and since there are $n^n$ endofunctions we obtain
$$E(z) = 1 + z\frac{d}{dz} T(z) = 1 + z T'(z).$$
But from the functional equation we get
$$T'(z) = \exp T(z) + z \exp T(z) T'(z) = \frac{T(z)}{z} + T(z) T'(z)$$
so that
$$T'(z) = \frac{1}{z} \frac{T(z)}{1-T(z)}.$$
This finally yields
$$E(z) = 1 + \frac{T(z)}{1-T(z)}$$
and hence $$Q(z) = \log\left(1+\frac{T(z)}{1-T(z)}\right).$$
This gives the sequence
$$1, 3, 17, 142, 1569, 21576, 355081, 6805296,
\\ 148869153, 3660215680,\ldots$$
which points us to OEIS A001865
where we learn that indeed we have the right exponential generating
function. Note that the formula for $E(z)$ also proves that
$\mathcal{E} = \textsc{SEQ}(\mathcal{T}).$
Now to extract coefficients from this for a closed formula we expand
the logarithm to get for the count $q_n$ the formula
$$n! [z^n] \sum_{k\ge 1} \frac{(-1)^{k+1}}{k}
\left(\frac{T(z)}{1-T(z)}\right)^k.$$
Observe that we can restrict this to $k\le n$ because the tree
function term starts at $z,$ getting
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\left(\frac{T(z)}{1-T(z)}\right)^k.$$
We still need the terms of the fraction in the tree function,
which can be done by Lagrange inversion. We have
$$[z^n] \left(\frac{T(z)}{1-T(z)}\right)^k
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\left(\frac{T(z)}{1-T(z)}\right)^k \; dz.$$
Put $T(z) = w$ so that $z=w/\exp(w)=w\exp(-w)$ and
$dz=(\exp(−w)−w\exp(−w))\; dw$ to get
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n+1))}{w^{n+1}}
\left(\frac{w}{1-w}\right)^k
(\exp(−w)−w\exp(−w))\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(wn)}{w^{n+1-k}}
\frac{1}{(1-w)^{k-1}} \; dw .$$
Extracting the residue at zero yields for $k\ge 2$
$$\sum_{q=0}^{n-k} \frac{n^{n-k-q}}{(n-k-q)!}
{q+k-2\choose k-2}$$
and for $k=1,$ $$\frac{n^{n-1}}{(n-1)!}.$$
Collecting these into one formula finally yields
$$n^n +
n! \sum_{k=2}^n \frac{(-1)^{k+1}}{k}
\sum_{q=0}^{n-k} \frac{n^{n-k-q}}{(n-k-q)!}
{q+k-2\choose k-2}.$$
This computation is closely related to the material at this
MSE link.
Addendum. I just noticed in one of the other posts that fixed points are not admitted in those endofunctions. The above material admits fixed points as in the discussion and the diagram at the Wikipedia entry.
Addendum Thu Dec 18 19:57:09 CET 2014.
The case when there are no fixed points goes as follows.
There are now $(n-1)^n$ endofunctions with no fixed points,
which gives $$E(z) = 1 + \sum_{n\ge 1} (n-1)^n \frac{z^n}{n!}.$$
Now observe that when we apply Lagrange inversion to
$$\frac{\exp(-T(z))}{1-T(z)}$$ we obtain
$$[z^n] \frac{\exp(-T(z))}{1-T(z)}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\frac{\exp(-T(z))}{1-T(z)}\; dz$$
which using the same substitution as before becomes
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n+1))}{w^{n+1}}
\frac{\exp(-w)}{1-w}
(\exp(−w)−w\exp(−w))\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n-1))}{w^{n+1}}\; dw
= \frac{(n-1)^n}{n!}.$$
But $\exp(-T(z)) = \frac{z}{T(z)}$ and we finally get for
$E(z)$ the closed form
$$\frac{z}{T(z)(1-T(z))}.$$
This gives for $Q(z)$ that
$$Q(z) = \log\left(\frac{z}{T(z)(1-T(z))}\right)$$
(note that $E(z)$ has a constant term in its expansion at zero which
is one) which produces the sequence
$$0, 1, 8, 78, 944, 13800, 237432, 4708144, 105822432, 2660215680,
\\ 73983185000, 2255828154624,\ldots $$
which points us to OEIS A000435, confirming
the result from the accepted answer. Note that the OEIS
says that this sequence was the first in the database, so we are
content to be referencing it in this computation.
To get a closed form re-write $Q(z)$ as follows:
$$Q(z) = \log\left(1 + \frac{z}{T(z)(1-T(z))} -1\right)$$
to get the formula
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\left(\frac{z}{T(z)(1-T(z))} -1\right)^k$$
This is
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\sum_{q=0}^k {k\choose q} (-1)^{k-q}
\left(\frac{z}{T(z)(1-T(z))}\right)^q.$$
We can drop the term for $q=0$ when $n\ge 1,$ getting
$$n! [z^n] \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\sum_{q=1}^k {k\choose q} (-1)^{k-q}
\left(\frac{z}{T(z)(1-T(z))}\right)^q.$$
Use Lagrange inversion to extract coefficients from the tree function
term.
$$[z^n] \left(\frac{z}{T(z)(1-T(z))}\right)^q
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
\left(\frac{z}{T(z)(1-T(z))}\right)^q
\; dz$$
which using the same substitution as before becomes
$$\frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n+1-q))}{w^{n+1-q}}
\left(\frac{1}{w(1-w)}\right)^q
(\exp(−w)−w\exp(−w))\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\epsilon} \frac{\exp(w(n-q))}{w^{n+1}}
\frac{1}{(1-w)^{q-1}} \; dw$$
Extracting coefficients we get
$$\sum_{p=0}^n \frac{(n-q)^{n-p}}{(n-p)!} {p+q-2\choose q-2}$$
when $q\ge 2$ and for $q=1$
$$\frac{(n-1)^n}{n!}.$$
Substituting this into the sum formula yields
$$n! \times \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\times k \times (-1)^{k-1} \frac{(n-1)^n}{n!}
\\ + n! \times \sum_{k=1}^n \frac{(-1)^{k+1}}{k}
\sum_{q=2}^k {k\choose q} (-1)^{k-q}
\sum_{p=0}^n \frac{(n-q)^{n-p}}{(n-p)!} {p+q-2\choose q-2}.$$
This simplifies to
$$n\times (n-1)^n
+ n! \times \sum_{k=2}^n \frac{(-1)^{k+1}}{k}
\sum_{q=2}^k {k\choose q} (-1)^{k-q}
\sum_{p=0}^n \frac{(n-q)^{n-p}}{(n-p)!} {p+q-2\choose q-2}.$$
This formula can be used to compute the count of these graphs for
large $n$ where the tree function formula would no longer be
practicable.
The sequence for $n=30$ to $n=34$ reads
38086159100543376291945674612050231296000000,
3117962569860399657478478640723143576082043800,
263711778692997479722657378560127779200642842624,
23019602620026625886784119896351926037410391377792,
2071846675499818842878197235287956993753027358752768
Additional observations.
The OEIS entry OEIS A000435
says that this sequence is the normalized total height of all labeled
rooted trees on $n$ nodes (sum of height of all nodes in all trees
scaled by $n$). The question arises on how to prove this.
The height parameter is represented by the bivariate generating
function $T(z, u)$ where $T(z, 1) = T(z)$ (the ordinary tree
function) and we have the functional equation
$$T(z, u) = z
+ z\frac{T(uz,u)^1}{1!}
+ z\frac{T(uz,u)^2}{2!}
+ z\frac{T(uz,u)^3}{3!}
+ z\frac{T(uz,u)^4}{4!}
+ \cdots$$
or
$$T(z, u) = z \exp T(uz, u).$$
The exponential generating function for the sum of the height of all
nodes of all rooted labeled trees by the number of nodes is given by
$$G(z) = \left.\frac{\partial}{\partial u} T(z, u)\right|_{u=1}.$$
We thus differentiate the functional equation, getting
$$G(z) = \left. z \exp T(uz, u)
\left(z \frac{\partial}{\partial z} T(z, u) +
\frac{\partial}{\partial u} T(z, u) \right)\right|_{u=1}$$
which becomes
$$G(z) = T(z) (z T'(z) + G(z))$$
so that
$$G(z) (1-T(z)) = z T(z) T'(z)$$
or
$$G(z) = \frac{z T(z)}{1-T(z)} T'(z).$$
We may substitute the expression for the derivative of the tree
function that we obtained earlier into this to get
$$G(z) = \frac{z T(z)}{1-T(z)}
\frac{1}{z} \frac{T(z)}{1-T(z)}
= \left(\frac{T(z)}{1-T(z)}\right)^2.$$
We have computed the exponential generating function of the total
height of all labelled trees on $n$ nodes, which gives the sequence
$$0, 2, 24, 312, 4720, 82800, 1662024, 37665152, 952401888,
\\ 26602156800, 813815035000 $$
which is OEIS A001864.
The normalized height is this sequence divided by $n.$ To verify that
this is indeed the same as the count of endofunctions with no fixed
point we must show that
$$z \frac{d}{dz} Q(z) = G(z)$$
i.e. that the endofunctions times $n$ give the total height or
alternatively, the total height divided by $n$ give the endofunctions.
But the left is
$$ z \frac{T(z)(1-T(z))}{z} \\ \times
\left(\frac{1}{T(z)(1-T(z))}
- z \frac{1-2T(z)}
{T(z)^2 (1-T(z))^2} T'(z) \right)$$
which gives
$$1- z \frac{1 - 2T(z)}{T(z)(1-T(z))} T'(z)
= 1- \frac{1 - 2T(z)}{T(z)(1-T(z))} \frac{T(z)}{1-T(z)}
= 1- \frac{1 - 2T(z)}{(1-T(z))^2}
\\ = \frac{1-2T(z)+T(z)^2 -(1- 2T(z))}{(1-T(z))^2}
= \left(\frac{T(z)}{1-T(z)}\right)^2,$$
thus concluding the proof.
