Closed Form for the Imaginary Part of $\text{Li}_3\Big(\frac{1+i}2\Big)$
If you consider a hypergeometric function to be a closed form, you can have the following result: $$\Im\left[\operatorname{Li}_3\left(\frac{1+i}2\right)\right]=\frac{\pi^3}{128}+\frac\pi{32}\ln^22+\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right).$$ And for the polylogarithm value that appears in another answer, you can have $$\Im\Big[\operatorname{Li}_3\left(1+i\right)\Big]=\frac{3\pi^3}{64}+\frac\pi{16}\ln^22-\frac14\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right).$$
Form here we know that $$\operatorname{Li}_3(z)-\operatorname{Li}_3\left(\frac{1}{z}\right)=-\frac{1}{6} \ln^3(z)-\frac{\pi\sqrt{-(z-1)^2}}{2(z-1)}\ln^2(z)+\frac{\pi^2}{3}\ln(z)\tag{1}.$$
If we put $z:=(1+i)/2$ into $(1)$ and get the imaginary part of it, we get
$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] - \Im \left[\operatorname{Li}_3(1-i)\right] = \frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2 2.$$
@Tunk-Fey said that $\displaystyle\Im[\operatorname{Li}_3(1-i)]=-\Im[\operatorname{Li}_3(1+i)]$, so it is also true, that
$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] + \Im \left[\operatorname{Li}_3(1+i)\right] = \frac{7\pi^3}{128}+\frac{3\pi}{32}\ln^2 2.$$
Sadly the only two exact complex value of $\operatorname{Li}_3$ function what I found is $\operatorname{Li}_3(\pm i).$ We could use it, but with known identities we coudn't break out of the prison of $\operatorname{Li}_3(1+i)$ or $\operatorname{Li}_3(1-i)$ with this approach.
I found nothing about it, but I think @Tunk-Fey's result is in general true, and I think that is true, that $\Im \operatorname{Li}_3(z)+\Im \operatorname{Li}_3(\overline{z}) = 0$ for all $z$ which have complex part. It isn't a good news, because using identities we get an unknown complex value of $\operatorname{Li}_3$ or a complex conjugate pair of the variable what we are looking for.
Of course there is relationship between polylogarithm function and generalized hypergeometric function. For $\operatorname{Li}_3$ we have $$\operatorname{Li}_3(z) = z \;_{4}F_{3} (1,1,\dots,1; \,2,2,\dots,2; \,z).$$
So we could write the problem also into the form
$$\Im\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] = \Im\left[\frac{1+i}{2}{_4F_3}\left(\begin{array}c\ 1,1,1,1\\2,2,2\end{array}\middle|\,\frac{1+i}{2}\right) \right].$$
I don't know how @Cleo transformed it into the form that is given above, but it is really nice, how the imaginary part is eliminated.
If somebody could give some more complex valued exact solutions of $\operatorname{Li}_3$ function, then maybe I could get something more. But for now, I'm also waiting for a solution.
By the way, if your result about the real part is correct, then we can get a really beautiful closed formula for $\Re[\operatorname{Li}_3(1 \pm i)]$ using the partial results of your problem.
If we put again $z:=(1+i)/2$ into $(1)$ and now we get the real part of it, we have
$$\Re\left[\operatorname{Li}_3\left(\frac{1+i}{2}\right)\right] - \Re \left[\operatorname{Li}_3(1-i)\right] = \frac{\ln^3 2}{48}-\frac{11\pi^2}{192}\ln 2.$$
Now if you're right, and
$$\Re\bigg[\text{Li}_3\bigg(\dfrac{1+i}2\bigg)\bigg]=\dfrac{\ln^32}{48}-\dfrac5{192}~\pi^2~\ln2+\dfrac{35}{64}~\zeta(3),$$
then we get for $\Re \left[\operatorname{Li}_3(1 \pm i)\right]$ the following.
$$\Re \left[\operatorname{Li}_3(1 \pm i)\right] = \frac{\pi^2}{32} \ln 2 + \frac{35}{64} \zeta(3).$$
This seems to me numerically correct, and via your problem I could solve a really interesting related diamond.
This is equivalent to finding a closed form for the following series
$$I :=\mathrm{Im} \left[\mathrm{Li}_3\left(\frac{1+i}{2}\right)\right] = \sum_{k=1}^{\infty} \frac{\sin \pi k/4}{2^{k/2}k^3}.$$
I was able to find the following expression, which seems to be numerically correct
$$I=\frac{7\pi^3}{256}+\frac{3\pi}{64}\log^22+\frac{1}{2} \mathrm{Im}\left[\mathrm{Li}_3\left(\frac{1}{2}+\frac{i}{2}\right)-\mathrm{Li}_3(1+i)\right].$$
Update: Taking the advice of V.Rosssetto into account, we are reduced to
$$I=\frac{7\pi^3}{128}+\frac{3\pi}{32}\log^22-\mathrm{Im} \; \mathrm{Li}_3(1+i).$$