Generalized Harmonic Number Summation $ \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}$

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Prove That $$ \sum_{n=1}^{\infty} \dfrac{(H_{n}^{(2)})^2}{2^n} = \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 $$

Notation : $ \displaystyle H_{n}^{(2)} = \sum_{r=1}^{n} \dfrac{1}{r^2}$

We can solve the above problem using the generating function $\displaystyle \sum_{n=1}^{\infty} (H_{n}^{(2)})^2 x^n $, but it gets rather tedious especially taking into account the indefinite polylogarithm integrals involved. Can we solve it using other methods like Euler Series Transform or properties of summation?