$$ \begin {aligned} &\int _{0}^{1}\int _{0}^{1}\frac {\ln ^{n}\frac 1x\ln (1-xy)}{1-x(1-y)}dxdy\\ &=\int _{0}^{1}\int _{0}^{1}\frac {\ln ^{n}\frac {1}x\ln (1-x+xy)}{1-xy}dxdy\\ &=\iint _{0< y< x< 1}\frac {\ln ^{n}\frac {1}x}x\frac {\ln (1-x+y)}{1-y}dxdy\\ &=-\int _{0}^{1}\left (\left .\frac {\ln ^{n+1}\frac {1}x}{n+1}\frac {\ln (1-x+y)}{1-y}\right |_{x=y}^1+\frac {1}{n+1}\int _{y}^{1}\frac {\ln ^{n+1}\frac {1}x}{(1-y)(1-x+y)}dx\right )dy\\ &=-\frac {1}{n+1}\iint _{0< y< x< 1}\frac {\ln ^{n+1}\frac {1}x}{2-x}\left (\frac {1}{1-y}+\frac {1}{1-x+y}\right )dydx\\ &=-\frac {1}{n+1}\int _{0}^{1}\frac {\ln ^{n+1}\frac {1}x}{2-x}\cdot 2\ln \frac {1}{1-x}dx\\ &=-\frac {2}{n+1}\int _{0}^{1}\frac {\ln ^{n+1}\frac {1}{1-x}\ln \frac {1}x}{1+x}dx\\ &=-\frac {2}{n+1}\cdot (n+1)!(-1)\sum _{0< r_1< \cdots < r_{n+2}}\frac {(-1)^{r_{n+2}-r_{n+1}}}{r_1\cdots r_{n+1}r_{n+2}^2}\\ &=2\cdot n!\zeta (\{1\}^n,\bar {1},\bar {2}) \end {aligned} $$

$$ \begin{align*} &\int_0^1\int_0^1\int_0^1\frac{\log^n(1-x(1-y(1-z)))}{1-xyz}dxdydz\\ =&\iiint_{0< z< y< x< 1}\frac{\log^n(1-x+y-z)}{1-z}\frac1{xy}dxdydz\\ =&\iiint_{0< z< y< x< 1}\frac{\log^n(1-y)}{1-z}\frac1{x(x-y+z)}dxdydz\\ =&\iint_{0< z< y< 1}\frac{\log^n(1-y)}{(1-z)(y-z)}\left[\log\left(1-\frac{y-z}x\right)\right]_y^1dydz\\ =&\iint_{0< z< y< 1}\frac{\log^n(1-y)}{(1-z)(y-z)}\log\left(\frac yz(1-y+z)\right)dydz\\ =&\iint_{0< y,z< 1}\frac{\log^n(1-y)}{(1-yz)(1-z)}\log\frac{1-y(1-z)}zdydz\\ =&\iint_{0< y,z< 1}\frac{\log^n(1-y)}{(1-y(1-z))z}\log\frac{1-yz}{1-z}dydz\\ =&-\iint_{0< y,z< 1}\frac{\log^n(1-y)}{(1-y(1-z))z}\int_y^1\frac{\partial}{\partial x}\log(1-xz)dxdydz\\ =&\iiint_{0< y< x< 1}\int_0^1\frac{\log^n(1-y)}{(1-y(1-z))(1-xz)}dzdxdy\\ =&-\iint_{0< y< x< 1}\frac{\log^n(1-y)}{1-(1-x)(1-y)}\left[\log\frac{1-xz}{1-y(1-z)}\right]_0^1 dxdy\\ =&-\iint_{0< y< x< 1}\frac{\log^n(1-y)}{1-(1-x)(1-y)}\log\left((1-x)(1-y)\right)dxdy\\ =&-\iint_{0< x< y< 1}\frac{\log^ny}{1-xy}\log xydxdy\\ =&-\int_0^1\frac{\log^n y}y\left[\text{Li}_2(1-xy)\right]_0^y dy\\ =&-\int_0^1\frac{\log^n y}y\left(\text{Li}_2(1-y^2)-\zeta(2)\right)dy\\ =&-\frac1{n+1}\left[\log^{n+1}y\left(\text{Li}_2(1-y^2)-\zeta(2)\right)\right]_0^1-\frac4{n+1}\int_0^1\frac{y\log y}{1-y^2}\log^{n+1}ydy\\ =&\frac1{(n+1)2^{n+1}}\int_0^1\frac{\log^{n+2}x}{1-x}dx\\ =&\frac1{(n+1)2^{n+1}}\sum_{k=1}^\infty\int_0^1 x^{k-1}\log^{n+2}xdx\\ =&\frac{(-1)^{n+2}}{(n+1)2^{n+1}}(n+2)!\sum_{k=1}^\infty\frac1{k^{n+3}}\\ =&\frac{(-1)^n\Gamma(n+3)}{(n+1)2^{n+1}}\zeta(n+3) \end{align*} $$