For the case of $n$ labeled nodes we find using PIE the closed form
$$D_n= \sum_{p=0}^n {n\choose p} (-1)^p 4^{n-p\choose 2}.$$ This will
produce $$0, 3, 54, 3861, 1028700, 1067510583, 4390552197234, \ldots$$
which points to OEIS A054545 where it
appears we have a match.
For the unlabeled case observe that the number $F_n$ of non-isomorphic
digraphs was computed at the following MSE
link. We then
obtain for the number of digraphs with no isolated nodes
$$D_n = F_n - F_{n-1}$$
which will produce
$$0, 2, 13, 202, 9390, 1531336, 880492496, 1792477159408,\ldots$$
again a match, this time with OEIS
A053598.
More data concerning PIE. The nodes of the poset for use with PIE
correspond to all subsets $P$ of vertices of the $n$ vertices and
represent labeled digraphs where the vertices in $P$, plus possibly
some other vertices, are isolated. The weight on the the digraphs
specified by $P$ is $(-1)^{|P|}.$ This means the digraphs with no
isolated vertices have weight one because they appear only in the node
that corresponds to $P=\emptyset.$ Digraphs with a set $Q$ of isolated
vertices where $|Q|\ge 1$ appear in all nodes $P\subseteq Q,$ for a
total weight of
$$\sum_{P\subseteq Q} (-1)^{|P|} =
\sum_{p=0}^{|Q|} {|Q|\choose p} (-1)^p = 0,$$
i.e. zero. The cardinality of the set of diagraphs corresponding to
node $P$ is $4^{n-|P|\choose 2}.$ We now sum the weights over all
digraphs, collecting the contributions from all nodes where they
appear. We already know that this will assign weight one to those
with no isolated vertices and zero otherwise, providing the desired
count. There are ${n\choose p}$ sets $P$ of $p=|P|$ nodes and there
are $4^{n-p\choose 2}$ digraphs at these nodes with weight $(-1)^p$,
concluding the derivation of the formula using PIE. Note that the
count is the same regardless of whether we iterate over all digraphs,
collecting the weights from the nodes or over all nodes, collecting
the weights of all digraphs.
