For future reference because we are using an EGF which may not be
admissible here. Recall that the species of set partitions is given by
$$\mathfrak{P}(\mathfrak{P}_{\ge 1}(\mathcal{Z})).$$
We seek to evaluate $${n\brace n-3}.$$
Now since these sets are not empty we must first put a labelled ball
into each of the $n-3$ slots. This leaves three labeled balls. We can
partition these as follows: $3$, $1+2$ and $1+1+1.$
The first case yields
$$\mathfrak{P}_{=4}(\mathcal{Z})
\mathfrak{P}_{=n-4}(\mathcal{Z}).$$
The second case yields
$$\mathfrak{P}_{=2}(\mathcal{Z})\mathfrak{P}_{=3}(\mathcal{Z})
\mathfrak{P}_{=n-5}(\mathcal{Z}).$$
The third case yields
$$\mathfrak{P}_{=3}(\mathfrak{P}_{=2}(\mathcal{Z}))
\mathfrak{P}_{=n-6}(\mathcal{Z}).$$
The term involving $n$ represents the singleton sets.
We thus get for the generating function
$$G(z) = \frac{z^4}{24}\frac{z^{n-4}}{(n-4)!}
+ \frac{z^2}{2}\frac{z^3}{6}\frac{z^{n-5}}{(n-5)!}
+ \frac{1}{6}
\left(\frac{z^2}{2}\right)^3 \frac{z^{n-6}}{(n-6)!}.$$
Extracting coefficients from this yields
$$n! [z^n] G(z)
= {n\choose 4} + 10 {n\choose 5} + 15 {n\choose 6}.$$
Addendum. We can treat the general case when we evaluate
$${n\brace n-k}$$
in terms of binomial coefficients
$${n\choose n-k-q} = {n\choose k+q}$$
where $1\le q\le k.$
Solving these for small $k$ points us to
OEIS A269939 (Ward numbers)
where we find the formula
$${n\brace n-k}
= \sum_{q=1}^k {n\choose n-k-q}
\sum_{m=0}^q {k+q\choose k+m} (-1)^{m+q}
{k+m\brace m}$$
which we now prove.
Recalling the bivariate generating function for set partitions
(Stirling numbers of the second kind) which is
$$G(z, u) = \exp(u(\exp(z)-1))$$
we get for the sum
$$\sum_{q=1}^k {n\choose n-k-q}
\\ \times \sum_{m=0}^q \frac{(k+q)!}{(k+m)!(q-m)!} (-1)^{m+q}
(k+m)! [z^{k+m}] \frac{(\exp(z)-1)^m}{m!}
\\ = n! \sum_{q=1}^k \frac{1}{(n-k-q)!} \frac{1}{q!}
\\ \times \sum_{m=0}^q {q\choose m} (-1)^{m+q}
[z^{k+m}] (\exp(z)-1)^m
\\ = \frac{n!}{(n-k)!} \sum_{q=1}^k {n-k\choose q}
\sum_{m=0}^q {q\choose m} (-1)^{m+q}
[z^{k+m}] (\exp(z)-1)^m.$$
Now introduce
$$[z^{k+m}] (\exp(z)-1)^m
= \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+m+1}} (\exp(z)-1)^m
\; dz$$
to get for the inner sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}}
\sum_{m=0}^q {q\choose m} (-1)^{q-m}
\frac{(\exp(z)-1)^m}{z^m}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}}
\left(\frac{\exp(z)-1}{z}-1\right)^q
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+q+1}}
\left(\exp(z)-z-1\right)^q
\; dz.$$
This yields for the outer sum
$$\frac{n!}{(n-k)!} \sum_{q=1}^k {n-k\choose q}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+q+1}}
\left(\exp(z)-z-1\right)^q
\; dz.$$
Now observe carefully that when $q\gt k$ then $2q\gt k+q$ and since
$\exp(z)-z-1$ starts at $z^2/2$ and hence $(\exp(z)-z-1)^q$ at
$z^{2q}/2^q$ the integral is zero in this case. Furthermore with
$k\ge 1$ we also get zero from the integral when $q=0.$ Therefore we
are justified in letting the sum range from zero to $n-k$ and we
obtain
$$\frac{n!}{(n-k)!}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}}
\sum_{q=0}^{n-k} {n-k\choose q} \frac{1}{z^q}
\left(\exp(z)-z-1\right)^q
\; dz
\\ = \frac{n!}{(n-k)!}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+1}}
\left(1 + \frac{\exp(z)-z-1}{z}\right)^{n-k}
\; dz
\\ = \frac{n!}{(n-k)!}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}}
\left(\exp(z)-1\right)^{n-k}
\; dz.$$
This is
$$n! [z^n] \frac{(\exp(z)-1)^{n-k}}{(n-k)!}
= {n\brace n-k}$$
and we are done.
Note that when $k\gt n-k$ we are also justified in lowering the
upper limit of the sum to $n-k$ because the segment being omitted is
zero due to the binomial coefficient ${n-k\choose q}.$
Alternate ending. We may use the formula for the outer sum to
obtain a combinatorial expression. Start from
$$\frac{n!}{(n-k)!} \sum_{q=1}^k {n-k\choose q}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+q+1}}
\left(\exp(z)-z-1\right)^q
\; dz
\\ = \sum_{q=1}^k \frac{n!}{(n-k-q)!}
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+q+1}}
\frac{\left(\exp(z)-z-1\right)^q}{q!}
\; dz
\\ = \sum_{q=1}^k {n\choose k+q}
\frac{(k+q)!}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{k+q+1}}
\frac{\left(\exp(z)-z-1\right)^q}{q!}
\; dz
\\ = \sum_{q=1}^k {n\choose k+q} {k+q\brace q}_{\ge 2}.$$
This is ${n\brace n-k}$ by inspection. Recall that we start by placing
a labeled singleton in each of the $n-k$ slots. The remaining $k$
items are divided into $q$ sets where $1\le q\le k.$ Each of these
joins a singleton, so in fact a configuration is completely determined
by the elements that are in a set containing at least two
elements. That's what the formula on the last line represents -- first
choose the $k+q$ items for those sets and combine that choice with a
partition of these items into $q$ sets of cardinality at least
two. (We have $k+q$ items because the singletons from the beginning
contribute $q$ of them.)
Remark. Having reached the end of this computation we see that the
variable in the integral was never substituted and rested inert. That
means we could have used the coefficient extractor notation throughout
without affecting the semantics of the argument.
