Let us look at the case $x=1$ in here. Note that the Nielsen generalized poly-logarithm at plus unity always reduces to single zeta values only. We know that from Compute an integral containing a product of powers of logarithms. for example.The only thing we need to do is to evaluate the three integrals. We do this in the usual way,i.e. by expanding the integrand in a multivariate series , integrating term by term and then using partial fraction decomposition to evaluate the sums. Now we state the results. We start from the easiest case and end with the hardest one. We have:
\begin{eqnarray}
&&\int\limits_0^1 \frac{[Li_3(\xi)]^2}{\xi} \cdot \frac{[\log(1/\xi)]^{n-4}}{(n-4)!} d\xi=\sum\limits_{\lambda_1\ge 1}\sum\limits_{\lambda_2\ge 1} \frac{1}{\lambda_1^3} \frac{1}{\lambda_2^3} \frac{1}{(\lambda_1+\lambda_2)^{n-3}}=\\
&&\sum\limits_{l_1=2}^3 \binom{n-1-l_1}{3-l_1} (-1)^{3-l_1} \zeta(l_1) \zeta(n-l_1+3) + \sum\limits_{l_1=2}^{n-3} \binom{n-1-l_1}{2} (-1)^3 \zeta(l_1,n-l_1+3) + \binom{n-2}{2} \left[ \zeta(n+2,1) + \zeta(n+3)\right]
\end{eqnarray}
The remaining two integrals require already a triple summation. We have:
\begin{eqnarray}
&&\int\limits_0^1 \frac{Li_0(\xi) Li_1(\xi) Li_2(\xi)}{\xi} \cdot \frac{[\log(1/\xi)]^{n-1}}{(n-1)!} d\xi=
\sum\limits_{\lambda_1\ge 1}\sum\limits_{\lambda_2\ge 1}\sum\limits_{\lambda_3\ge 1} \frac{1}{\lambda_1^0} \frac{1}{\lambda_2^1} \frac{1}{\lambda_3^2} \frac{1}{(\lambda_1+\lambda_2+\lambda_3)^n}=\\
&&-\sum\limits_{i=1}^{n-1} \zeta(n+1-i,i,2)+\\
&&\zeta(2) \cdot\left[\zeta(n,1)+\zeta(n+1)\right]-\left[\zeta(2,n,1)+\zeta(2,n+1)\right]-\left[\zeta(n+2,1)+\zeta(n+3)\right]
\end{eqnarray}
and
\begin{eqnarray}
&&\int\limits_0^1 \frac{[Li_1(\xi)]^2}{\xi} \cdot Li_l(\xi) \cdot \frac{[\log(1/\xi)]^{n-l}}{(n-l)!} d\xi=\sum\limits_{\lambda_1\ge 1}\sum\limits_{\lambda_2\ge 1}\sum\limits_{\lambda_3\ge 1} \frac{1}{\lambda_1^1} \frac{1}{\lambda_2^1} \frac{1}{\lambda_3^l} \frac{1}{(\lambda_1+\lambda_2+\lambda_3)^{n-l+1}}=\\
&&-\sum\limits_{i=1}^{n-l} \sum\limits_{j=1}^l \zeta(n-l+2-i,i+j,l+1-j) \cdot\left(1_{j<l} + 2 \delta_{j,l}\right)+\\
&&\left.\sum\limits_{j=1}^l \left(\zeta(i+j+1,l+1-j)+\zeta(i+j,l+2-j) +\zeta(i+j,l+1-j,1)+\zeta(i+j,1,l+1-j)\right)\cdot\left(1_{j<l} + 2 \delta_{j,l}\right)\right|_{i=n-l+1}
\end{eqnarray}
Note 1: All the bi-variate zeta functions which we have above are in principle reducable to single zeta values. We know that from Calculating alternating Euler sums of odd powers for example.
Note 2: The question whether the triple-variate zeta values above are too reducable to single zeta values is open for the time being. It would be nice to know under what circumstances this is the case.
