Integrations by parts of the recursive definition of $\;\operatorname{Li}_n(x)$ : $\;\displaystyle\operatorname{Li}_{n+1}(x)=\int\frac {\operatorname{Li}_{n}(x)}x\,dx\;$ allowed Lewin (in his $1981$ reference book "Polylogaritms and associated functions") to write :
$$\tag{7.62}\operatorname{Li}_4(x)=\log(x)\operatorname{Li}_3(x)-\frac 12\log^2(x)\operatorname{Li}_2(x)-\frac 16\log^3(x)\log(1-x)-\frac 16\int_0^x\frac{\log^3(y)}{1-y}\,dy$$
for $\; x:=1-e^{it}\;$ this becomes
$$\tag{7.66}\operatorname{Li}_4(1-e^{it})=\log(1-e^{it})\operatorname{Li}_3(1-i)-\frac 12\log^2(1-i)\operatorname{Li}_2(1-e^{it})-\frac {it}6\log^3(1-e^{it})\\+\frac i6\int_0^{t}\log^3(1-e^{iv})\,dv$$
We may rewrite the last integral as $\;\displaystyle\int_0^{t}\left(\frac i2(v-\pi)+\log\left(2\sin\frac v2\right)\right)^3\,dv\;$ to expand it using binomials in terms of generalized log-sine integrals $\;\displaystyle\operatorname{Ls}_j^{(k)}(t):=-\int_0^t v^k\,\left(\log\left(2\sin\frac v2\right)\right)^{j-k-1}\,dv$.
After quite some rewriting and reduction Lewin obtained his equation $(7.68)$ for the real part :
\begin{align}
&\Re\operatorname{Li}_4\left(1-e^{it}\right)=\frac 14\operatorname{Ls}_4^{(1)}\left(t\right)-\frac t4\operatorname{Ls}_3\left(t\right)+\frac {t^2}8\log^2\left(2\sin\frac t2\right)+\frac{\operatorname{Li}_3(1)-\operatorname{Cl}_3(t)}2\log\left(2\sin\frac t2\right)-\frac{t^4}{192}\\
&\text{giving for $t=\frac {\pi}2\;$ since $\;\displaystyle\operatorname{Cl}_3\left(\frac {\pi}2\right)=-\frac{3}{32}\zeta(3)$ :}\\
\tag{1}&\Re\operatorname{Li}_4\left(1-i\right)=\frac 14\operatorname{Ls}_4^{(1)}\left(\frac {\pi}2\right)-\frac {\pi}8\operatorname{Ls}_3\left(\frac {\pi}2\right)+\frac{\pi^2}{32}\log^2\left(\sqrt{2}\right)+\frac {35}{64}\zeta(3)\log\left(\sqrt{2}\right)-\frac{\pi^4}{3072}\\
\end{align}
But the two log-sine terms disappear using the first of the $(A.14)$ relations :
$$\operatorname{Ls}_{4}^{(1)}\left(\tfrac{\pi}{2}\right)-\tfrac{\pi}{2} \operatorname{Ls}_{3}\left(\tfrac{\pi}{2}\right) = -\tfrac{5}{96} \tag{2}\log^4(2)
+ \tfrac{5}{16} \zeta(2) \log^2(2) - \tfrac{35}{32} \zeta(3) \log(2)
+ \tfrac{125}{32} \zeta(4) - \tfrac{5}{4} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)\\
$$
as provided by Davydychev and Kalmykov in the appendix of their paper "New results for the epsilon-expansion of certain one-, two- and three-loop Feynman diagrams" (from this SE answer detailing the notations used here)
$(1)$ then becomes :
\begin{align}
\Re\operatorname{Li}_4\left(1-i\right)&=\frac 14\left[-\tfrac{5}{96} \log^4(2)
+ \tfrac{5}{16} \zeta(2) \log^2(2) - \tfrac{35}{32} \zeta(3) \log(2)
+ \tfrac{125}{32} \zeta(4) - \tfrac{5}{4} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)\right]+\frac{\pi^2}{32}\log^2\left(\sqrt{2}\right)+\frac {35}{64}\zeta(3)\log\left(\sqrt{2}\right)-\frac{\pi^4}{3072}\\
&=- \frac{5}{16} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)-\frac{5}{384} \log^4(2) + \frac{5}{64} \zeta(2) \log^2(2)
+ \frac{125}{128} \zeta(4) +\frac{\pi^2}{128}\log^2\left(2\right)-\frac{\pi^4}{3072}\\
\tag{3}\Re\operatorname{Li}_4\left(1-i\right)&=- \frac{5}{16} \operatorname{Li}_{4}\left(\tfrac{1}{2}\right)-\frac{5}{384} \log^4(2)+ \frac{97}{9216}\pi^4+ \frac{\pi^2}{48}\log^2\left(2\right)\\
\end{align}
Which is exactly your result ($\Re\operatorname{Li}_4\left(1+i\right)=\Re\operatorname{Li}_4\left(1-i\right)$ of course).
The only problem is that the relations $(A.14)$ from the paper of Davydychev and Kalmykov is followed by the words (much sweeter for physicists than for mathematicians...) :
$\qquad$"All relations $(A.9)–(A.14)$ have been obtained using the PSLQ procedure".
So that $(A.14)$ may or not have been proved since $2001$ (the paper was updated in $2017$). Anyway the remaining problem could be to prove the expression $(2)$ with the LHS given by $\;\displaystyle \int_0^{\frac{\pi}2} \left(\frac {\pi}2-t\right)\log^2\left(2\sin\frac t2\right)\,dt$
