TITLE: Known exact values of the $\operatorname{Li}_3$ function
QUESTION [13 upvotes]: We know some exact values of the trilogarithm $\operatorname{Li}_3$ function.
Known real analytic values for $\operatorname{Li}_3$:

$\operatorname{Li}_3(-1)=-\frac{3}{4} \zeta(3)$
$\operatorname{Li}_3(0)=0$
$\operatorname{Li}_3\left( \frac{1}{2} \right) = \frac{7}{8} \zeta(3) - \frac{1}{12} \pi^2 \ln 2 + \frac{1}{6} \ln^3 2$
$\operatorname{Li}_3(1) = \zeta(3),$ where $\zeta(3)$ is the Apéry's constant
$\operatorname{Li}_3\left(\phi^{-2}\right)=\frac{4}{5} \zeta(3) + \frac{2}{3} (\ln \phi)^3 - \frac{2}{15} \pi^2 \ln \phi,$ where $\phi$ is the golden ratio.

Using identities for the list above we could also get:

$\operatorname{Li}_3(2) = \frac{\pi^2}{4} \ln 2 + \frac{7}{8} \zeta(3) - \frac{\pi}{2} \ln^2(2) \cdot i,$ or we could write into this alternate form.
$\operatorname{Li}_3\left(\phi^2\right) = \frac{4}{5} \zeta(3) - \frac{2}{3} \ln^3 \phi + \frac{8\pi^2}{15}\ln \phi - 2\pi\ln^2(\phi) \cdot i,$ or there is an alternate form here.

We know even less about complex argumented values:

$\operatorname{Li}_3(i)=-\frac{3}{32}\zeta(3) +\frac{\pi^3}{32} i$
$\operatorname{Li}_3(-i)=-\frac{3}{32}\zeta(3) -\frac{\pi^3}{32} i.$

There are some partial result for complex cases:

$\Im\left[ \operatorname{Li}_3 \left( \frac{1 \pm i}{\sqrt{2}} \right) \right] = \pm \frac{7\pi^3}{256}$, and there is an expression for the real part in term of derivatives of digamma function,
other related specific values around the unit circle like this or this, etc.

Lucian said the following in this question:

$\Re\left[\text{Li}_3\left(\dfrac{1+i}2\right)\right]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3).$

While working on Lucian's problem I was able to specify this one

$\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$

Than I got the idea to write a question, maybe someone could give us some more specific values of the trilogarithm function.

REPLY [5 votes]: It appears that
$$
\Im\left[\mathrm{Li}_3\left((-1)^{\frac{1}{2^n}}\right) \right]=\frac{(2^n-1)(2^{n+1}-1)}{3\cdot2^{3n+2}}\pi^3
$$
but that's by experimentation.