Here is an attempta way. Start from the exact form of your integral expression (i.e. supply the "Dirichlet factor" $\frac{\Gamma(\alpha_1+\ldots+\alpha_n)}{\Gamma(\alpha_1)\cdots\Gamma(\alpha_n)})$.
(1) Writing $\frac{1}{(x_1+\ldots+x_n)^{\alpha_1+\ldots+\alpha_n}}=\int_0^{\infty}\frac{ t^{\alpha_1+\ldots+\alpha_n-1}}{\Gamma(\alpha_1+\ldots+\alpha_n)}\,e^{-t(x_1+\ldots+x_n)}\,dt$ and changing the order of integration (and denoting by $p$$p_{n,N}$ the distribution of $\tau(N)(P)$$\tau(N)(P)=:(M_{1,N},\ldots,M_{n,N})$) leads to $$p(m_1,\ldots,m_n)=\sum_{j=1}^n \int_0^\infty \frac{m_j^{\alpha_j}t^{\alpha_j-1}}{\Gamma(\alpha_j)}\prod_{i\neq j}\big (R_{\alpha_i}(m_it)-R_{\alpha_i}((m_i+1)t)\big) e^{-m_jt}\,dt$$$$p_{n,N}(m_1,\ldots,m_n)=\sum_{j=1}^n \int_0^\infty \frac{m_j^{\alpha_j}t^{\alpha_j-1}}{\Gamma(\alpha_j)}\prod_{i\neq j}\big (R_{\alpha_i}(m_it)-R_{\alpha_i}((m_i+1)t)\big) e^{-m_jt}\,dt$$ where $R_{\alpha} (t):=\int_t^\infty \frac{y^{\alpha-1}}{\Gamma(\alpha)} e^{-y}\,dy$ (and $m_1+\ldots+m_n=N$).
(2) For $\alpha=k$ (a positive integer) $$R_k(t)=q_k(t) e^{-t} \mbox{ with } q_k(t)=\sum_{i=0}^{k-1}\frac{t^i}{i!}\;\;.$$ If all $\alpha_i$ are all integers the integrals above therefore lead to explicit rational expressions for the probabilities $p(m_1,\ldots,m_n)$. For the case $\alpha_1=\ldots=\alpha_{n-1}=1, \alpha_n=k$ (positive integer) one gets \begin{align*} p(m_1,\ldots,m_n)= &\frac{m_n^k}{(k-1)!}\int_0^\infty t^{k-1} e^{-Nt} (1-e^{-t})^{n-1}\,dt\\& + (N-m_n)\int_0^\infty \big(q_k(m_nt)-e^{-t}q_k((m_n+1)t)\big) e^{-Nt}(1-e^{-t})^{n-2}\,dt\end{align*}\begin{align*} p_{n,N}(m_1,\ldots,m_n)= &\frac{m_n^k}{(k-1)!}\int_0^\infty t^{k-1} e^{-Nt} (1-e^{-t})^{n-1}\,dt\\& + (N-m_n)\int_0^\infty \big(q_k(m_nt)-e^{-t}q_k((m_n+1)t)\big) e^{-Nt}(1-e^{-t})^{n-2}\,dt\end{align*} ADDED: a fully explicit solution for the case $\alpha_1=\ldots=\alpha_{n-1}=1, \alpha_n=k$ (and this can easily be workedpositive integer).
(3) In the sequel $n\geq 2$, and $k$ is fixed and suppressed from the notation. Working out explicitly usingthe integrals above gives $\Gamma$-integrals(for any nonnegative integers $m_1,\ldots,m_{n-1},m$ with $m_1+\ldots +m_n+m=N$). \begin{align*} p_{n,N}(m_1,\ldots,m_{n-1},m)=I_{n-1,N}(m) + J_{n-2,N}(m) \end{align*} where \begin{align*} I_{n,N}(m)&=m^k\sum_{j=0}^{n}(-1)^j {n \choose j}\frac{1}{(N+j)^k}\\ J_{n,N}(m)&=\sum_{j=0}^{n} (-1)^j {n \choose j} \frac{N-m}{N-m+j}\Big(\big(\frac{m+1}{N+j+1}\big)^k-\big(\frac{m}{N+j}\big)^k\Big)\\ \end{align*}
Since the joint distribution $p(m_1,\ldots,m_n)$$p_{n,N}(m_1,\ldots,m_{n-1},m)$ depends on $m_1,\ldots,m_{n-1}$ only through $m_1+\ldots+m_{n-1}=N-m_n$ the$m_1+\ldots+m_{n-1}=N-m$ (we will from now on simply write $p_{n,N}(m)$, and) the distribution of$f_{n,N}(m):=\mathbb{P}(M_{n,N}=m)$ of the last coordinate $M_n$$M_{n,N}$ is proportional simply given by \begin{align*} f_{n,N}(m)= {N-m+n-2 \choose n-2} \cdot p_{n,N}(m) \end{align*} (since $ {N-m+n-2 \choose n-2}$ is the number of $n-1$-tuples of nonnegative integers summing to $p$$N-M$). I assume
(4) Examples
(a) For $k=1$ both integrals above are the same, and we get \begin{align*} p_{n,N}(m)=N\int_0^\infty e^{-Nt} (1-e^{-t})^{n-1}\,dt=N\, B(N-1,n-1)=\frac{N!\,(n-1)!}{(N+n-1)!}\;\;, \end{align*} confirming that $\tau(N)(P)$ is uniformly distributed on the discrete simplex (but haven't checkedas was already noted above).
(b) For $n=2$ we find $I_{1,N}=m^k\big(\frac{1}{N^k}-\frac{1}{(N+1)^k}\big), J_{0,N}=(\frac{m+1}{N+1})^k-(\frac{m}{N})^k$ so that computer algebra
\begin{align*} f_{2,N}(m)=\frac{(m+1)^k-m^k}{(N+1)^k}=p_{2,N}(m),\;\;\;\;m=0,\ldots,N \end{align*}
and similar computations give for $n=3$
\begin{align*} f_{3,N}(m)=& \Big((m+1)^k-m^k\Big)\Big(\frac{N+2}{(N+1)^k}-\frac{N+1}{(N+2)^k}\Big) \\&-\Big((m+1)^{k+1}+m^{k+1}\Big)\Big(\frac{1}{(N+1)^k}-\frac{1}{(N+2)^k}\Big) \end{align*} (5) The distribution of $M_{n,N}$ for arbritrary $n$. Although the above description of the distribution of $\tau(N)(P)$ resp. $M_{n,N}$ is (in principle) complete, it is not very transparent. A surprisingly simple description can now findbe obtained if one uses finite differences. Let \begin{align*} \delta_n(N):=\sum_{j=0}^n{n \choose j} \frac{(-1)^j}{(N+j)^k}\end{align*} (so that $(-1)^n \delta_n(N)$ is the $n$-th forward difference of $1/x^k$ at $N$). We then have
Lemma \begin{align*} (1)&\;\;{N-m+n \choose n}J(n,N)=\sum_{j=0}^n {N-m-1+j \choose j}\Big((m+1)^k\,\delta_j(N+1) -m^k \delta_j(N)\Big)\;\;\\ (2)&\;\;f_{n+2,N}(m)=\sum_{j=0}^n\Big[(m+1)^k{N-(m+1)+j \choose j}-m^k{N-m+j \choose j}\Big]\;\delta_j(N+1)\;\;\;. \end{align*} Proof (Sketch) (1): use (show) the equality $$\sum_{j=i}^n {N-m-1+j \choose j}{ j \choose i}=\frac{N-m}{N-m+i}{N-m+n \choose n} {n \choose i}\;\;\;.$$ (2): the coefficients of $(m+1)^k$ follow directly from (1). The coefficients of $m^k$ follow by using $I(n,N)=m^k\delta_{n+1}(N)$, $\delta_{n}(N)=\delta_{n-1}(N)-\delta_{n-1}(N+1)$ and the addition rule for binomial cofficients.;- End Proof
(6) The expectation of $M_{n,N}$
From (2) of the lemma $\mathbb{P}(M_{n+2,N}\leq m)=\sum_{j=0}^n (m+1)^k{N-(m+1)+j \choose j}\;\delta_j(N+1)$ and thus \begin{align*} \mathbb{E}(M_{n+2,N})&=\sum_{m=0}^{N-1} \mathbb{P}(M_{n+2,N}>m)\\ &=\sum_{m=0}^{N-1} \big(1-\mathbb{P}(M_{n+2,N}\leq m)\big)\\ &=N -\sum_{j=0}^n\Big[\sum_{m=1}^N m^k{N-m+j \choose j}\Big]\;\delta_j(N+1) \end{align*} The sums $\sum_{m=1}^N m^k{N-m+j \choose j}$ can be expressed in various ways.
Denoting the Stirling numbers of the second kind by ${k \brace d}$, the falling factorials by $(m)_d$ and using $m^k=\sum_{d=1}^k { k \brace d} (m)_d$ and $\sum_{m=1}^N (m)_d {N-m+j \choose j}=d!{N+j+1 \choose d+j+1}$ gives \begin{align*} \sum_{m=1}^N m^k {N-m+j \choose j}=\sum_{d=1}^k d!\,{ k \brace d}{N+j+1 \choose d+j+1} \end{align*} Plugging this in above gives an explicit expressionsformula for the expectation of. Alternatively one could could expand the binomial cofficient as a polynomial in $M_n$$m$, and use power sums/Faulhabers formula (as has been done in the solution below).