Is there a closed form expression for \begin{align} e^{\Large\frac{i\pi}3} \text{Li}_{2}\left( \frac{e^{\Large\frac{i\pi}3} }{2}\right) + e^{-\Large\frac{i\pi}3} \text{Li}_{2}\left( \frac{e^{-\Large\frac{i\pi}3} }{2}\right) \end{align} in terms of known constants such as: $\pi$, $\ln 2$, $\ln 3$, $e$, $\gamma$, etc.
The expression to be calculated can be placed into series form and is equivalent to calculating the series \begin{align} \sum_{n=1}^{\infty} \frac{ \cos\left( \frac{(n+1)\pi}{3} \right) }{2^{n-1} \ n^{2}} \end{align} in closed form.
Letting $z_{\pm}=\frac{e^{\pm\frac{i\pi}{3}}}{2}$, we find,
$$\frac{z_{\pm}}{z_{\pm}-1}=\mp\frac{i}{\sqrt{3}},$$
and
$$\ln{\left(1-z_{\pm}\right)}=\frac12\ln{\left(\frac34\right)}\mp i\frac{\pi}{6},\\ \implies \ln^2{\left(1-z_{\pm}\right)}=\frac14\ln^2{\left(\frac43\right)}-\frac{\pi^2}{36}\pm i \frac{\pi}{6}\ln{\left(\frac43\right)}.$$
Then using Landen's dilogarithm identity,
$$\begin{align} \operatorname{Li}_{2}{\left(z_{\pm}\right)} &=-\operatorname{Li}_{2}{\left(\frac{z_{\pm}}{z_{\pm}-1}\right)}-\frac12\ln^2{\left(1-z_{\pm}\right)}\\ &=-\operatorname{Li}_{2}{\left(\mp\frac{i}{\sqrt{3}}\right)}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)}\pm i \frac{\pi}{12}\ln{\left(\frac34\right)}. \end{align}$$
The real and imaginary components of $\operatorname{Li}_{2}{\left(z_{+}\right)}$ are then,
$$\Re{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}=-\Re{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)}\\ =-\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}+\frac{\pi^2}{72}-\frac18\ln^2{\left(\frac43\right)},$$
and,
$$\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}=-\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}+\frac{\pi}{12}\ln{\left(\frac34\right)}.$$
The term $\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]}$ may be further simplified with the aid of the dilogarithmic identity,
$$\operatorname{Li}_{2}{\left(z\right)}-\operatorname{Li}_{2}{\left(-z\right)}+\operatorname{Li}_{2}{\left(\frac{1-z}{1+z}\right)}-\operatorname{Li}_{2}{\left(-\frac{1-z}{1+z}\right)}=\frac{\pi^2}{4}+\ln{\left(z\right)}\ln{\left(\frac{1+z}{1-z}\right)}.$$
Setting $z=-\frac{i}{\sqrt{3}}$,
$$\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}-\operatorname{Li}_{2}{\left(\frac{i}{\sqrt{3}}\right)}+\operatorname{Li}_{2}{\left(e^{\frac{i\pi}{3}}\right)}-\operatorname{Li}_{2}{\left(-e^{\frac{i\pi}{3}}\right)} = \frac{\pi^2}{12}+i\frac{\pi}{6}\ln{(3)}.$$
$$\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}-\operatorname{Li}_{2}{\left(\frac{i}{\sqrt{3}}\right)} = i\frac{\pi}{6}\ln{(3)}+i\frac{5\pi^2}{9\sqrt{3}}-i\frac{5}{6\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$
$$\Im{\left[\operatorname{Li}_{2}{\left(-\frac{i}{\sqrt{3}}\right)}\right]} = \frac{\pi}{12}\ln{(3)}+\frac{5\pi^2}{18\sqrt{3}}-\frac{5}{12\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$
$$\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]} = -\frac{\pi}{6}\ln{(2)}-\frac{5\pi^2}{18\sqrt{3}}+\frac{5}{12\sqrt{3}}\psi^{(1)}{\left(\frac13\right)}$$
The quantity of interest is then:
$$\begin{align} A &=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2z_{-}\operatorname{Li}_{2}{\left(z_{-}\right)}\\ &=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2\,\overline{z_{+}}\operatorname{Li}_{2}{\left(\overline{z_{+}}\right)}\\ &=2z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}+2\,\overline{z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}}\\ &=4\Re{\left[z_{+}\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}\\ &=\Re{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}-\sqrt{3}\,\Im{\left[\operatorname{Li}_{2}{\left(z_{+}\right)}\right]}\\ &=-\frac14\operatorname{Li}_{2}{\left(-\frac13\right)}+\frac{7\pi^2}{24}-\frac18\ln^2{\left(\frac34\right)}+\frac{\sqrt{3}\,\pi}{6}\ln{(2)}-\frac{5}{12}\psi^{(1)}{\left(\frac13\right)}.\\ \end{align}$$
I don't think the sum of dilogarithm trigamma terms in the final line can be simplified to a combination of the basic constants suggested.