Major Edit: This is almost a completely new response rather than an edit. The previous version of this response was extremely long and clumsy, and ultimately failed to even yield a definite final result. This new and improved response is much more streamlined and does include a definite final value.
Evaluation of integral $I_3$:
The triple integral defining $I_3$ can in principle be integrated in any order, but integrating in "alphabetical order" (i.e., with the integral over $x$ as the outermost one, and the integral over $z$ as the innermost) is probably the best way to go and is the order used in the first step below. Next, we rescale the integral over $z$ via the substitution $t=(xy)\,z$; after that, we also rescale the integral over $y$ via the substitution $u=(x)\,y$. Now, instead of evaluating the integrals from innermost-to-outermost, note that our integral is now in a form that lends itself very well to integration by parts with respect to $x$. The result is a sum of two double integrals:
$$\begin{align}
I_3
&=\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\ln{(1-x)}\ln{(1-xy)}\ln{(1-xyz)}\,\mathrm{d}x\mathrm{d}y\mathrm{d}z\\
&=\int_{0}^{1}\mathrm{d}x\,\ln{(1-x)}\int_{0}^{1}\mathrm{d}y\,\ln{(1-xy)} \int_{0}^{1}\mathrm{d}z\,\ln{(1-xyz)}\\
&=\int_{0}^{1}\mathrm{d}x \frac{\ln{(1-x)}}{x} \int_{0}^{x}\mathrm{d}u \frac{\ln{(1-u)}}{u}\int_{0}^{u}\mathrm{d}t\,\ln{(1-t)}\\
&=-\operatorname{Li}_2{(1)}\int_{0}^{1}\mathrm{d}u \frac{\ln{(1-u)}}{u}\int_{0}^{u}\mathrm{d}t\,\ln{(1-t)}\\
&~~~~~+\int_{0}^{1}\mathrm{d}x\frac{\operatorname{Li}_2{(x)}\ln{(1-x)}}{x}\int_{0}^{x}\mathrm{d}t\,\ln{(1-t)}\\
&=:J+K.
\end{align}$$
Using the evaluations of $J$ and $K$ below, we arrive at a final value for $I_3$:
$$\begin{align}
I_3
&=J+K\\
&=\left[-3\zeta{(2)}+2\zeta{(3)}\zeta{(2)}\right]+\left[\zeta{(5)}+6\zeta{(4)}+6\zeta{(3)}+3\zeta{(2)}-2\zeta{(3)}\zeta{(2)}-15\right]\\
&=\zeta{(5)}+6\zeta{(4)}+6\zeta{(3)}-15\\
&=\zeta{(5)}+6\zeta{(3)}+\frac{\pi^4}{15}-15\\
&=-0.2567 9142 3632 2352\dots .
\end{align}$$
$$I_3=\color{blue}{\zeta{(5)}+6\zeta{(3)}+\frac{\pi^4}{15}-15}.$$
Evaluation of integral $J$:
First we state without proof the following three anti-derivatives:
$$\int\mathrm{d}u\ln{(1-u)}=(u-1)\ln{(1-u)}-u+constant;$$
$$\int\mathrm{d}u\ln^2{(1-u)}=(u-1)\left(\ln^2{(1-u)}-2\ln{(1-u)}+2\right)+constant;$$
$$\int\mathrm{d}u\frac{\ln^2{(1-u)}}{u}=-2\operatorname{Li}_3{(1-u)}+2\operatorname{Li}_2{(1-u)}\ln{(1-u)}+\ln{(u)}\ln^2{(1-u)}+constant.$$
They may each be easily verified by differentiating the right-hand-sides, or checked using WolframAlpha. Once obtained, the integral $J$ may be calculated directly from these three anti-derivatives:
$$\begin{align}
J
&=-\operatorname{Li}_2{(1)}\int_{0}^{1}\mathrm{d}u \frac{\ln{(1-u)}}{u}\int_{0}^{u}\mathrm{d}t\,\ln{(1-t)}\\
&=-\operatorname{Li}_2{(1)}\int_{0}^{1}\mathrm{d}u \frac{\ln{(1-u)}}{u}\left[(u-1)\ln{(1-u)}-u\right]\\
&=\operatorname{Li}_2{(1)}\int_{0}^{1}\mathrm{d}u \left[\ln{(1-u)}-\ln^2{(1-u)}+\frac{\ln^2{(1-u)}}{u}\right]\\
&=\operatorname{Li}_2{(1)} \left[-1-2+2\zeta{(3)}\right]\\
&=-3\zeta{(2)}+2\zeta{(3)}\zeta{(2)}.
\end{align}$$
Evaluation of integral $K$:
$$\begin{align}
K
&=\int_{0}^{1}\mathrm{d}x\frac{\operatorname{Li}_2{(x)}\ln{(1-x)}}{x}\int_{0}^{x}\mathrm{d}t\,\ln{(1-t)}\\
&=\int_{0}^{1}\mathrm{d}x\frac{\operatorname{Li}_2{(x)}\ln{(1-x)}}{x} \left[(x-1)\ln{(1-x)}-x\right]\\
&=\int_{0}^{1}\mathrm{d}x \operatorname{Li}_2{(x)}\ln{(1-x)} \left[-1+\ln{(1-x)}-\frac{\ln{(1-x)}}{x}\right]\\
&=\int_{0}^{1}\mathrm{d}x \operatorname{Li}_2{(x)} \left[-\ln{(1-x)}+\ln^2{(1-x)}-\frac{\ln^2{(1-x)}}{x}\right]\\
&=-\int_{0}^{1}\mathrm{d}x \ln{(1-x)}\operatorname{Li}_2{(x)} + \int_{0}^{1}\mathrm{d}x \ln^2{(1-x)}\operatorname{Li}_2{(x)} - \int_{0}^{1}\mathrm{d}x \frac{\ln^2{(1-x)} \operatorname{Li}_2{(x)}}{x}\\
&=K_1+K_2+K_3\\
&=\left[2\zeta{(3)}+\zeta{(2)}-3\right]+\left[6\zeta{(4)}+4\zeta{(3)}+2\zeta{(2)}-12\right]+\left[\zeta{(5)}-2\zeta{(3)}\zeta{(2)}\right]\\
&=\zeta{(5)}+6\zeta{(4)}+6\zeta{(3)}+3\zeta{(2)}-2\zeta{(3)}\zeta{(2)}-15.
\end{align}$$
The evaluations of $K_1$ and $K_2$ can easily be found by first using a CAS to find the anti-derivatives. Finally, $K_3$ is calculated below.
Evaluation of integral $K_3$:
$$\begin{align}
K_3
&=-\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(1-x)} \operatorname{Li}_2{(x)}}{x}\\
&=-\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(1-x)} \left[\zeta{(2)}-\ln{(1-x)}\ln{(x)}-\operatorname{Li}_2{(1-x)}\right]}{x}\\
&=-\zeta{(2)}\int_{0}^{1}\mathrm{d}x\frac{\ln^2{(1-x)}}{x} + \int_{0}^{1}\mathrm{d}x \frac{\ln^3{(1-x)}\ln{(x)}}{x}+\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(1-x)}\operatorname{Li}_2{(1-x)}}{x}.
\end{align}$$
The first integral in the last line above has already been calculated as part of the evaluation of integral $J$. The evaluations of the second and third integrals can be found in the responses to this question and this question, respectively.
$$\begin{align}
K_3
&=-\zeta{(2)}\int_{0}^{1}\mathrm{d}x\frac{\ln^2{(1-x)}}{x} + \int_{0}^{1}\mathrm{d}x \frac{\ln^3{(1-x)}\ln{(x)}}{x}+\int_{0}^{1}\mathrm{d}x \frac{\ln^2{(1-x)}\operatorname{Li}_2{(1-x)}}{x}\\
&=-2\zeta{(3)}\zeta{(2)}+\left[12\zeta{(5)}-6\zeta{(3)}\zeta{(2)}\right]+\left[-11\zeta{(5)}+6\zeta{(3)}\zeta{(2)}\right]\\
&=\zeta{(5)}-2\zeta{(3)}\zeta{(2)}.
\end{align}$$