Hypergeometric series for $\mathrm{Cl}_2(\pi/3)$

Well, you have already shown that $\text{Cl}_2\left(\frac{\pi}{3}\right)$ is proportional to $$ \sum_{n\geq 0}\frac{1}{(3n+1)^2}-\sum_{n\geq 0}\frac{1}{(3n+2)^2} =\int_{0}^{1}\frac{-\log(x)\,dx}{1+x+x^2}$$ i.e. to $$ \phantom{}_3 F_2\left(\tfrac{1}{3},\tfrac{1}{3},1;\tfrac{4}{3},\tfrac{4}{3};1\right)-\phantom{}_3 F_2\left(\tfrac{2}{3},\tfrac{2}{3},1;\tfrac{5}{3},\tfrac{5}{3};1\right) $$ and you may apply Thomae's theorem to derive many equivalent representations.

This is Gieseking's constant and it does have a hypergeometric form,

$$\begin{aligned} \rm{Cl}_2\left(\frac{\pi}{3}\right) &= \frac{3\sqrt3}4\left(\sum_{n=0}^\infty\frac{1}{(3n+1)^2}-\sum_{n=0}^{\infty}\frac{1}{(3n+2)^2}\right)\\ &= \,_3 F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;\tfrac14\right)\\ &= 1.0149416\dots \end{aligned}$$

---

Added edit: More generally,

$$\rm{Ls}_1\big(\tfrac\pi3\big) = \int_0^{\pi/3}\log^0\left(2\sin\frac{t}2\right)dt=0!\,(-1)^0\;_2 F_1\left(\tfrac12,\tfrac12;\tfrac32;\tfrac14\right)=\tfrac\pi3$$

$$\rm{Ls}_2\big(\tfrac\pi3\big) = \int_0^{\pi/3}\log^1\left(2\sin\frac{t}2\right)dt=1!\,(-1)^1\;_3 F_2\left(\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32;\tfrac14\right)=-\rm{Cl}_2\left(\tfrac{\pi}{3}\right)$$

$$\rm{Ls}_3\big(\tfrac\pi3\big) = \int_0^{\pi/3}\log^2\left(2\sin\frac{t}2\right)dt=2!\,(-1)^2\;_4 F_3\left(\tfrac12,\tfrac12,\tfrac12,\tfrac12;\tfrac32,\tfrac32,\tfrac32;\tfrac14\right) = \frac{7\pi^3}{216}$$

and closed-forms of $\rm{Ls}_n\big(\tfrac\pi3\big)$ for $n \leq 8$ (skipping $n=7$) are given here. Or,

$$\begin{aligned} \rm{Ls}_{n}\big(\tfrac\pi3\big) &= \int_0^{\pi/3}\ln^{n-1}\left(2\sin\frac{t}2\right)dt\\ &= -(n-1)!\,(-1)^n\sum_{k=0}^\infty\frac{1}{k!}\frac{\left(\big(\frac12\big)_k\right)^{n+1}}{\left(\big(\frac32\big)_k\right)^{n}}\frac1{4^k}\\ &= -(n-1)!\,(-1)^n\sum_{k=0}^\infty\frac{1}{k!}\frac{\big((2k-1)!!\big)^{n+1}}{\big((2k+1)!!\big)^{n}}\frac1{8^k}\\ &= -(n-1)!\,(-1)^n\sum_{k=0}^\infty\frac{\binom{2k}{k}}{(2k+1)^{n}} \frac1{16^k} \end{aligned}$$

where $(x)_k$ is the Pochhammer symbol, and the special case $\rm{Ls}_{2}\big(\tfrac\pi3\big)=-\rm{Cl}_{2}\big(\tfrac\pi3\big)$.