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: Here is a solution of a related problem, followed by a recommendation for the original problem. It would be much simpler if your sum did not have the $n$ in $(n+|m|)!\,$. In that case, we could look at the related sum
$$t_n=\sum_{m\in {\mathfrak{M} }_n}\frac{|m|!}{m!}\prod_{k=1}^n(k+1)^{-m_k}.$$
The sum for the $t$'s comes from a product of exponential generating functions. Because of the factor of $(k+1)^{-m_k}$ in $t_n$ and the term $k\,m_k$ in ${\mathfrak{M} }_n$, we must look at the series
$$1+\frac{\left(\frac{x^k}{k+1}\right)^1}{1!} +\frac{\left(\frac{x^k}{k+1}\right)^2}{2!} +\frac{\left(\frac{x^k}{k+1}\right)^3}{3!} +\cdots=\exp\left(\frac{x^k}{k+1}\right).$$
From multiplying these exponential generating functions, we get
$$t_n=\left[\frac{x^n}{n!}\right]\prod_{k\ge1}\exp\left(\frac{x^k}{k+1}\right).$$
This product turns out to have a nice closed form:
\begin{eqnarray*}
% \nonumber to remove numbering (before each equation)
  \prod_{k\ge1}\exp\left(\frac{x^k}{k+1}\right) &=& \exp\left(\sum_{k\ge1}\frac{x^k}{k+1}\right) \\
    &=& \exp\left(\frac1x\biggl(\log\bigl(\frac1{1-x}\bigr)-x\bigr)\right) \\
    &=& (1-x)^{-x}/\mathrm{e} .
\end{eqnarray*}
The smallest singularity of $(1-x)^{-x}$ is at 1, so a crude approximation would be
$$[x^n](1-x)^{-x}\approx1^n=1$$
and
$$t_n=\left[\frac{x^n}{n!}\right](1-x)^{-x}/\mathrm{e}\approx n!/\mathrm{e}.$$
Certainly, a finer analysis of the singularity of $(1-x)^{-x}$ would give a better approximation and perhaps produce the power $\alpha$ you're seeking.
Now, back to the original problem. It's always the case that $|m|\le n$, so a rough bound on $s_n$ would be $$s_n\le(2n)!t_n\le(2n)!\, .$$ This bound is worse than the hope you expressed, but perhaps good enough for your eventual purposes or perhaps a start for finer analysis.
