Computation of a series NOTATIONS.
Let $n\in\mathbb{N}$. We define the sets $\mathfrak{M}_{0}:=\emptyset$ and
\begin{align}
\mathfrak{M}_{n}&:=\left\{m=\left(m_{1},m_{2},\ldots,m_{n}\right)\in\mathbb{N}^{n}\mid1m_{1}+2m_{2}+\ldots+nm_{n}=n\right\}&\forall n\geq1
\end{align}
and we use the notations:
\begin{align}
m!&:=m_{1}!m_{2}!\ldots m_{n}!,&|m|&:=m_{1}+m_{2}+\ldots+m_{n}.
\end{align}
QUESTION.
I want to evaluate or just bound with respect to $n$ the series
\begin{align}
S_{n}&:=\sum_{m\in\mathfrak{M}_{n}}\frac{\left(n+\left|m\right|\right)!}{m!}\ \prod_{k=1}^{n}\left(k+1\right)^{-m_{k}}.
\end{align}
My hope is that $S_{n}\leq n!n^{\alpha}$ with $\alpha$ independant of $n$.
BACKGROUND.
In order to build an analytic extension from a given real-analytic function, I had to use the Faà di Bruno's formula for a composition (see for example https://en.wikipedia.org/wiki/Faà_di_Bruno%27s_formula). After some elementary computations, my problem boils down to show the convergence of
\begin{align}
\sum_{n=0}^{+\infty}\frac{x^{n+1}}{(n+1)!}\sum_{m\in\mathfrak{M}_{n}}\frac{\left(n+\left|m\right|\right)!}{m!}\ \prod_{k=1}^{n}\left(k+1\right)^{-m_{k}}
\end{align}
where $x\in\mathbb{C}$ is such that the complex modulus $|x|$ can be taken as small as desired (in particular, we can choose $|x|<\mathrm{e}^{-1}$ to kill any $n^{\alpha}$ term from the bound on $S_{n}$).
SOME WORK.
It is clear that we have to to understand the sets $\mathfrak{M}_{n}$ in order to go on (whence the tag "combinatorics"). So I tried to see what were these sets:


*

*for $n=2$ :
\begin{array}{cc}
2&0\\
0&1
\end{array}

*for $n=3$ :
\begin{array}{ccc}
3&0&0\\
1&1&0\\
0&0&1
\end{array}

*for $n=4$ :
\begin{array}{cccc}
4&0&0&0\\
2&1&0&0\\
1&0&1&0\\
0&2&0&0\\
0&0&0&1\\
\end{array}

*for $n=5$ :
\begin{array}{ccccc}
5&0&0&0&0\\
3&1&0&0&0\\
2&0&1&0&0\\
1&0&0&1&0\\
1&2&0&0&0\\
0&0&0&0&1\\
0&1&1&0&0\\
\end{array}


Above, each line corresponds to an multiindex $m$, and the $k$-th column is the coefficient $m_{k}$. We see for example that the cardinal of $\mathfrak{M}_{n}$ becomes strictly greater than $n$ if $n\geq5$. Also, because I wanted to reorder the set of summation in $S_{n}$ into a the set of all multiindices $m$ such that $|m|=j$ for $1\leq j\leq n$, I tried to count given $j$ the number of $m$ such that $|m|=j$; when $n=10$, I counted $8$ multiindices $m$ with length $|m|=4$, so that this number can be greater than $n/2$. Another remark is that the number of multiindices $m$ such that $|m|=j$ becomes larger if $j$ is "about" $n/2$ - don't ask me what "about" means here, I just tried some example and saw this phenomenon.
 A: Using $\frac1{k+1}<\frac1k$ and the notation $\sum im_i=\lambda\vdash n$ (equiv. $\lambda\in\mathfrak{M}_n$), begin estimating
\begin{align}
S_{n} 
<&\sum_{m\in\mathfrak{M}_{n}}\frac{\left(n+\left|m\right|\right)!}{m!}\ \prod_{k=1}^{n}k^{-m_{k}}
=\sum_{\lambda\vdash n}\frac{n!}{m!\prod k^{m_k}}\frac{(n+\vert m\vert)!}{n!}.
\end{align}
Note that $\frac{n!}{m!\prod k^{m_k}}$ is the number of permutations $\pi\in \mathfrak{S}_n$ having cycle type $(m_1,\dots,m_n)$ and $\vert m\vert=\kappa(\pi)=\#$ of cycles in $\pi$. In light of this,
\begin{align}
S_n<\sum_{\pi\in\mathfrak{S}_n}\frac{(n+\kappa(\pi))!}{n!}
=\sum_{j=1}^nc(n,j)\frac{(n+j)!}{n!};
\end{align}
where $c(n,j)$ are the (unsigned) Stirling numbers of the first kind. On the other hand, the numbers $a_n:==\sum_{j=1}^nc(n,j)(n+1)\cdots(n+j)$ are listed on OEIS A052819 and there Vaclav Kotesovec provideda growth estimate
$$a_n \qquad \sim \qquad (1+r)^n(2+r)^n\left(\frac{n}e\right)^n.$$
Combining this with Stirling approximation $n!\sim \sqrt{2\pi n}\left(\frac{n}e\right)^n$, we obtain
$$\frac1{n+1}\frac{S_n}{n!} \qquad \sim \qquad \frac{\sqrt{2\pi n}}{n+1}
(1+r)^n(2+r)^n$$
where $r=0.794862961852611133\cdot$. Hence, the desired convergence of your series can be achieved form small $x$.
REMARK. The above estimates can be improved although it is not necessary for your purpose. For example, there is an expected number of cycles in a permutation (regarding $\kappa(\pi)$) given by 
$$H_n=1+\frac12+\frac13+\cdots+\frac1n\sim \log n.$$
A: Your $S_n$ starts with $1,1,5,41,469,6889,123605,2620169,64074901,1775623081,54989743445,...$; this seems to be A032188 on OEIS (number of labeled circular-rooted trees with $n$ leaves). The link contains lots of information if so. In particular, asymptotics given there is $S_{n-1}\sim\frac{n^{n-1}}{2e^n(1-\log(2))^{n-\frac12}}$
