$\newcommand{\Z}{\mathbb Z}\renewcommand{\b}{\binom}$Here we will express the sum in question as an ordinary integral of an ordinary sum, which is much easier to analyze than the original expression -- which was a sum over an $m$-dimensional integer simplex, involving multinomial coefficients.
We will also analyze the simpler expression in the case when $m\to\infty$ and $n\ge cm\ln m$ for a real constant $c>1$. Then the resulting upper bound is optimal up to a constant factor.
For positive integers $n$ and $m$, let
\begin{equation*}
K^0:=\{k=(k_1,\dots,k_m)\in\Z_+^m\colon k_1+\dots+k_m=n\},
\end{equation*}
where $\Z_+:=\{0,1,\dots\}$, and
\begin{equation*}
K:=\{k\in K^0\colon k_1\ge1,\dots,k_m\ge1\}.
\end{equation*}
Then, by the exchangeability of the $k_i$'s, the sum in question is
\begin{equation*}
S_{n,m}:=m\sum_{k\in K}\b nk\frac1{k_m}, \tag{0}\label{0}
\end{equation*}
where
\begin{equation*}
\b nk:=\b n{k_1,\dots,k_m}=\frac{n!}{k_1!\dots,k_m!}.
\end{equation*}
So,
\begin{equation*}
S_{n,m}=m\int_0^1\frac{dx}x\,S(K), \tag{1}\label{1}
\end{equation*}
where
\begin{equation*}
S(J):=\sum_{k\in J}\b nk x^{k_m}
\end{equation*}
for all $J\in P([m])$, where in turn $P([m])$ is the power set of the set $[m]:=\{1,\dots,m\}$.
For $J\in P([m])$, let also
\begin{equation*}
K_J:=\{k\in K^0\colon k_j=0\ \forall j\in J\},
\end{equation*}
so that $K_\emptyset=K^0$ and $K_{[m]}=\emptyset$ (since $n\ge1$). So, by the inclusion-exclusion principle,
\begin{equation*}
S(K)=\sum_{j=0}^{m-1}(-1)^j\sum_{|J|=j}S(K_J), \tag{2}\label{2}
\end{equation*}
where $\sum_{|J|=j}$ denotes the summation over all $J\in P([m])$ of cardinality $|J|=j$.
Next, for each $J\in P([m])$ with $|J|=j$ we have the following:
If $m\in J$, then $S(K_J)=(m-j)^n$.
If $m\notin J$, then $S(K_J)=(x+m-j-1)^n$.
Also, there are $\b{m-1}{j-1}$ sets $J\in P([m])$ with $|J|=j$ such that $m\in J$; and there are $\b{m-1}{j}$ sets $J\in P([m])$ with $|J|=j$ such that $m\notin J$. So, by \eqref{2},
\begin{align*}
S(K)&=\sum_{j=0}^{m-1}(-1)^j\Big[\b{m-1}{j-1}(m-j)^n+\b{m-1}{j}(x+m-j-1)^n\Big] \\
&=\sum_{j=0}^{m-1}(-1)^j\b{m-1}{j}\big[(x+m-j-1)^n-(m-j-1)^n\big];
\end{align*}
here we took into account that $\b{m-1}{-1}=0$ and $m-m=0$ and $n\ge1$. So, in view of \eqref{1},
\begin{equation*}
S_{n,m}=m\sum_{j=0}^{m-1}(-1)^j\b{m-1}{j}I_{n,m}, \tag{3}\label{3}
\end{equation*}
where
\begin{align*}
I_{n,m}&:=\int_0^1\frac{dx}x\,\big[(x+m-j-1)^n-(m-j-1)^n\big] \\
&:=\int_0^1 dx\,\ln\frac1x\;n(x+m-j-1)^{n-1},
\end{align*}
via integration by parts. So, by \eqref{3},
\begin{equation*}
S_{n,m}=m\int_0^1 dx\,\ln\frac1x\;\sum_{j=0}^{m-1}(-1)^j\b{m-1}{j}n(x+m-j-1)^{n-1}. \tag{4}\label{4}
\end{equation*}
The latter expression for $S_{n,m}$, which is an ordinary integral of an ordinary sum, is much easier to analyze than the original expression \eqref{0} for $S_{n,m}$ (which was a sum over an $m$-dimensional integer simplex, involving multinomial coefficients).
Specifics of analysis of the expression for $S_{n,m}$ in \eqref{4} depend on how $m$ varies with $n$.
For instance, suppose that
\begin{equation*}
n\ge m\ln m, \tag{5}\label{5}
\end{equation*}
which seems a natural assumption.
Then for
\begin{equation*}
a_j:=a_{n,m;j}(x):=\b{m-1}{j}n(x+m-j-1)^{n-1}
\end{equation*}
and $x\in(0,1)$ and $j\in\{0,\dots,m-2\}$ we have
\begin{equation*}
r_j:=\frac{a_{j+1}}{a_j}=\Big(1-\frac1{x+m-j-1}\Big)^{n-1}\frac{m-j-1}{j+1} \\
\le\Big(1-\frac1m\Big)^{n-1}(m-1)\le m e^{-n/m}\le1.
\end{equation*}
So, the $a_j$'s are nonincreasing in $j\in\{0,\dots,m-1\}$.
So, by \eqref{4},
\begin{align*}
S_{n,m}&=m\int_0^1 dx\,\ln\frac1x\;\sum_{j=0}^{m-1}(-1)^j a_{n,m;j}(x) \\
&\le m\int_0^1 dx\,\ln\frac1x\;a_{n,m;0}(x) \\
&=m\int_0^1 dx\,\ln\frac1x\;n(x+m-1)^{n-1}.
\tag{6}\label{6}
\end{align*}
To upper-bound the latter integral, take any $b\in(0,1)$.
Since $\ln\frac1x$ is decreasing and $n(x+m-1)^{n-1}$ is increasing in $x\in(0,1)$, by the Chebyshev integral inequality we have
\begin{align*}
&\int_0^b dx\,\ln\frac1x\;n(x+m-1)^{n-1} \\
& \le\frac1b\,\int_0^b dx\,\ln\frac1x\;\int_0^b dx\,n(x+m-1)^{n-1} \\
& =\ln\frac1b\;[(b+m-1)^{n}-(m-1)^{n}] \\
& \le\ln\frac1b\;(b+m-1)^{n} \\
& \le\ln\frac1b\;m^n\Big(1+\frac{b-1}m\Big)^{n} \\
&\le\ln\frac1b\;m^n e^{-(1-b)n/m}.
\tag{7}\label{7}
\end{align*}
Since $\frac{\ln(1/x)}{1-x}$ is decreasing in $x\in(0,1)$, with
\begin{equation*}
C_b:=\frac{\ln(1/b)}{1-b}
\end{equation*}
we get
\begin{align*}
&\int_b^1 dx\,\ln\frac1x\;n(x+m-1)^{n-1} \\
& \le C_b\int_b^1 dx\,(1-x)\;n(x+m-1)^{n-1} \\
& \le C_b\int_0^1 dx\,(1-x)\;n(x+m-1)^{n-1} \\
& = C_b\frac mn[m^n-(m-1)^n-n(m-1)^{n-1}]\le C_b\frac{m^{n+1}}n.
\tag{8}\label{8}
\end{align*}
Collecting \eqref{6}, \eqref{7}, and \eqref{8}, we get
\begin{align*}
S_{n,m}&\le B_{n,m}:=m^{n+1}\Big(\frac{C_b}n+\ln\frac1b\;e^{-(1-b)n/m}\Big).
\tag{9}\label{9}
\end{align*}
If we now strengthen condition \eqref{5} to
\begin{equation*}
n\ge cm\ln m \tag{5a}\label{5a}
\end{equation*}
for any real constant $c>\frac1{1-b}$, then for the bound $B_{n,m}$ in \eqref{9} we will have
\begin{equation*}
B_{n,m}\sim C_b \frac{m^{n+1}}n
\end{equation*}
as $m\to\infty$.
This may be compared with the following trivial bound on $S_{n,m}$: in view of \eqref{0},
\begin{equation*}
S_{n,m}\le m\sum_{k\in K^0}\b nk=m^{n+1}. \tag{10}\label{10}
\end{equation*}
So, given \eqref{5a}, the bound $B_{n,m}$ in \eqref{9} improves the trivial bound in \eqref{10} by a factor $\asymp n$.
Following the lines of the above reasoning and noting that $\ln\frac1x>1-x$ for $x\in(0,1)$, one can see that this factor cannot be further improved, at least given \eqref{5a}.
Note also that, again assuming \eqref{5a}, the bound $B_{n,m}$ in \eqref{9} is better than the bound conjectured at the end of Brendan McKay's answer by a factor $\asymp m$.