Is the relation $P(n) \sim \frac{1}{2^n}$ already known? Apologies in advance if there is a violation of rules/laws here, as I am not a mathematician. 
$$
\begin{align}
\lim_{n\to\infty} \left( \frac{\pi^{n}}{\zeta(n)}P(n) \right)^{\frac{1}{n}} &= \frac{\pi}{2}
\\
\\
\lim_{n\to\infty} \left( \pi^{n}P(n) \right)^{\frac{1}{n}} &= \frac{\pi}{2}
\hspace{2cm}\lim_{n\to\infty} \zeta(n) = 1
\\
\\
\lim_{n\to\infty} \pi P(n)^{\frac{1}{n}} &= \frac{\pi}{2}
\\
\\
\lim_{n\to\infty} P(n)^{\frac{1}{n}} &= \frac{1}{2}
\\
\\
P(n) &\sim \frac{1}{2^n}
\end{align}
$$
Where $P(n)$ is the prime zeta function
$$P(n) = \sum_p \frac{1}{p^n}$$
Numerical calculations (from WolframAlpha) suggest this is true:
\begin{align*}
P(500) &\approx 3.054936363499604682051979393213617699789402740572326663... × 10^{-151} \\
2^{-500} &\approx 3.054936363499604682051979393213617699789402740572326663... × 10^{-151}\end{align*}

This is not an answer but it is too long for a comment.
I suppose that you could be interested by this paper where the author proposes $$\frac 1{P(n)}=2^n-\big(\frac 43\big)^n+\big(\frac 89\big)^n-\big(\frac 45\big)^n-\big(\frac {16}{27}\big)^n-\big(\frac 47\big)^n+2\big(\frac {8}{15}\big)^n+\big(\frac {32}{81}\big)^n+\cdots$$
For $P(10)$, the expansion (truncated to the  terms given here) gives an error $\approx -6.35 \times 10^{-12}$ and $P(10)\approx 0.000993604$ is already very close to $2^{-10} \approx 0.000976563$.
More empirical : generating the exact values of $P(n)$ for $10\leq n \leq 100$ and performing a nonlinear regression, we can get $$P(n)\approx 0.500741^n$$ the standard error on the tuned parameter being $0.000021$.
