It can be seen that: $$\sum_{n=1}^\infty n^{-2}=\frac{\pi^2}{6},\sum_{n=2}^\infty \frac{(-1)^n}{π(n)}=1$$ Where π(x) is the prime counting function with $π(0\le x<2)$=0 hence the starting index at 2. As there are two examples of convergence, I will add them together for more results. Note that each sum term converges separately. Sum approximation. You can think of it as the prime pi basel function. I will use many identities from here and Number Theory functions. The sums converge slowly:
$$\mathrm{P=\sum_{x=2}^\infty \left(\frac1{π^2(x)}+\frac1{π\left(x^2\right)}\right)=ln\prod_{x=2}^\infty e^{\frac{1}{\pi^2(x)}+\frac{1}{\pi\left(x^2\right)}}=4.71…}$$
$$\mathrm{\sum_{x=2}^\infty \frac1{\pi\big(x^2\big)}=1.86…, \sum_{x=2}^\infty\frac 1{\pi^2(x)}=2.84…}$$
Partial sums for P:
Summand graph:
It is possible to get bounds for $x\ge 2$ the summand as both bounds are positive. Graph of bounds:
$$\mathrm{\frac{x}{ln(x)+2}\le\pi(x)\le\frac{x\,ln(4)}{ln(x)}\implies \frac{log_4^2(x)+2log_4(x)}{x^2}\le\pi^{-2}(x)+\frac1{\pi\left(x^2\right)}\le\frac{ln^2(x)+6ln(x)+6}{x^2}}$$
This gives $2.236…\le \text P\le 6.532…$
I will add more forms. How would one evaluate P? An exact answer is needed. You can also find an integral representation of P. Please correct me and give me feedback!
With a few manipulation of the series we can prove that
$$\sum_{k=2}^\infty\frac{1}{\pi^2(k)} = \sum_{n=1}^\infty\frac{p_{n+1}-p_n}{n^2} = \sum_{n=2}^\infty p_n\frac{2n-1}{n^2(n-1)^2}-2$$
This series have a slightly better rate of convergence, but still looks far from a closed form (which I doubt exists).
The series $\displaystyle\sum_{k=2}^\infty\frac{1}{\pi(k^2)}$ seems much more difficult to handle.