Is $ \prod_{i=1}^{\infty} \frac{1}{(1-\frac{1}{2p_i^3})} $ equivalent to this sum of partitions?

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$$\prod_{i=1}^{\infty} \frac{1}{(1-\frac{1}{2 p_i^3})} = \sum_{n=0}^{\infty} 2^{-n} \sum_{k=0}^{\infty} (\frac{1}{2} ; \frac{1}{2})_k \: \frac{Par(n-\Omega(k+1)+k, k)}{(k+1)^3} $$

where $p_i$ is the $i$th prime and $ \Omega(n), (x;x)_k , Par(n,k) $ are the Prime Omega function, Q-Pochhammer symbol, and the Partition Triangle, respectively.

This is only conjectured and Mathematica gives a value of $1.0937396 \dots$ for both.

Edit: Expanding the LHS yields $$ \frac{\zeta^{\frac{1}{2}}(3)}{{\zeta^{\frac{1}{8}}(6)}\zeta^{\frac{1}{8}}(9) \zeta^{\frac{3}{64}}(12) \zeta^{\frac{3}{32}}(15) } \cdots $$ though I'm not sure if that helps working out the RHS.