We know that: $$\sum_{n=2}^\infty (\zeta(n)-1)=1$$ but what about with the Prime Zeta function $\text P(s)$?:
$$\sum_{n=2}^\infty \text P(n)=0.77315666904979…$$
Now interchange the sum with the prime numbers $ p_k$
$$\sum_{n=2}^\infty \sum_{m=1}^\infty \frac1{p_m^n}= \sum_{m=1}^\infty\sum_{n=2}^\infty \frac1{p_m^n}=\sum_{n=1}^\infty \frac1{p_n(p_n-1)}$$
which gives the same decimal. There are also many prime-related constants. Another form given by @reuns is with the Euler Phi and Mobius functions :
$$\sum_{n=2}^\infty \text P(n)=\sum_{n=2}^\infty \frac{\phi(n)\ln(\zeta(n))}n-\sum_{n=2}^\infty \frac{\mu(n)\ln(\zeta(n))}n= \sum_{n=2}^\infty \frac{\phi(n)\ln(\zeta(n))}n-\text C= \gamma-\text B_1+ \sum_{n=2}^\infty \frac{\phi(n)\ln(\zeta(n))}n $$
where $\text P(s)=-\ln(\epsilon)+\text C+O(\epsilon),\epsilon>0$ as seen in Formulas 3 throught 5 here on MathWorld and $\text C=\text B_1-\gamma$ with the Merten’s constant $\text B_1$ and Euler Mascheroni constant. It also gives the same decimal. The Euler phi sum is similar to @Steven Clark’s formula ($14$) here.
If one read further in the Merten’s constant article, then the amount of prime factors average deviation constant $\text B_2$ appears giving:
$$\sum_{n=2}^\infty \text P(n)=\sum_{p\text{ prime}}\frac1{p(p-1)}=\text B_2-\text B_1= 0.77315666904979… $$
Now that a “closed form” has been found, are there any alternate forms of the $0.77315666904979…\,$ constant in terms of special functions, integrals, manipulated sums et cetera?
This answer contains more information based on chapter 22 of "An Introduction to the Theory of Numbers" by Hardy and Wright which hopefully provides a bit more insight.
Let
$$C(x)=\sum\limits_{p\le x} \frac{\log(p)}{p}\tag{1}$$
where $p\in\mathbb{P}$ and
$$\tau(x)=C(x)-\log(x)=O(1)\tag{2}$$
then
$$\sum\limits_{p\le x} \frac{1}{p}=\frac{C(x)}{\log (x)}+\int\limits_2^x\frac{C(t)}{t \log^2(t)}\,dt=1+\frac{\tau (x)}{\log (x)}+\int_2^x \frac{1}{t \log(t)}\,dt+\int_2^x \frac{\tau (t)}{t \log ^2(t)} \, dt$$ $$=\log(\log(x))+B_1+E(x)\tag{3}$$
where
$$B_1=1-\log(\log(2))+\int_2^\infty \frac{\tau(t)}{t \log^2(t)}\,dt\tag{4}$$
and
$$E(x)=\frac{\tau(x)}{\log(x)}-\int\limits_x^\infty \frac{\tau(t)}{t \log^2(t)}\,dt=O\left(\frac{1}{\log(x)}\right)\tag{5}$$
so
$$\sum\limits_{p\le x} \frac{1}{p}=\log(\log(x))+B_1+o(1)\tag{6}$$
Also
$$B_1=\gamma+\sum\limits_p \left(\log\left(1-\frac{1}{p}\right)+\frac{1}{p}\right)\tag{7}$$
so
$$\sum_{n=2}^\infty P(n)=\sum\limits_{n=2}^\infty \frac{\phi(n)}{n} \log(\zeta(n))+\gamma-B_1\tag{8}=\sum\limits_{n=2}^\infty \frac{\phi(n)}{n} \log(\zeta(n))-\sum\limits_p \left(\log\left(1-\frac{1}{p}\right)+\frac{1}{p}\right)$$
A couple of other relationships are
$$\sum\limits_{n=1}^x \nu(n)=x \log(\log(x))+B_1 x+o(x)\tag{9}$$
$$\sum\limits_{n=1}^x \Omega(n)=x \log(\log(x))+B_2 x+o(x)\tag{10}$$
where $\nu(n)$ and $\Omega(n)$ are the number of distinct and non-distinct primes dividing $n$.