What does $\sum_{n=2}^{\infty} \ln(\zeta(n))$ converge to?

Question

Introduction

I was playing around with the zeta function when I thought of this series:

$$\sum_{n=2}^{\infty} \ln(\zeta(n))$$

According to WolframAlpha this sum converges and gives the following approximation:

$$\sum_{n=2}^{\infty} \ln(\zeta(n))\approx 0.83067$$

Questions

1. Is this a "known" series from before? I find this series quite cool and I thought I would find it somewhere but I didn't.
2. WolframAlpha says it confirmed the convergence by the root test, can anybody confirm this?
3. What's the closed-form of what the series converges to (exact value)?

Extra: My attempts

I tried to solve the 3rd question, but I don't have much ideas. An attempt I made was to note that:

\begin{align} \sum_{n=2}^{\infty} \ln(\zeta(n)) &= \sum_{n=2}^{\infty} \ln(\prod_p \frac{1}{1-p^{-n}}) \\\\ & = \sum_{n=2}^{\infty} \sum_p \ln(\frac{1}{1-p^{-n}})) \\\\ & = -\sum_{n=2}^{\infty} \sum_p \ln(1-p^{-n})) \end{align}

...but it didn't get me anywhere. Wikipedia also lists some integrals that are equal to $\ln(\zeta(n))$, but I'm not that familiar with them. Here's an example:

$$\displaystyle \ln \zeta (s)=s\int _{0}^{\infty }{\frac {\pi (x)}{x(x^{s}-1)}}\,\mathrm {d} x$$ And also: $$\displaystyle \ln \zeta (s)=s\int _{0}^{\infty }J(x)x^{-s-1}\,\mathrm {d} x$$ where $J(x)$ is the Riemann prime-counting function. I doubt that these are going to help, but I'll just put them here.

I'm not experienced in evaluating infinite series so I don't have anymore ideas. It would be highly appreciated if some of you could shed a little bit of light on this question!

Thanks in advance!

EDIT 1:

As Enver noted, this question is quite similar because my series is equal to $\ln(v)$ in the other question.

EDIT 2: Using basically the same method as in this question, we obtain (using one of the equations above):

\begin{align} \sum_{n=2}^{\infty} \ln(\zeta(n)) &= \sum_{n=2}^{\infty} n\int _{0}^{\infty }J(x)x^{-n-1}\,\mathrm {d} \\\\ & = \int _{0}^{\infty}J(x)(\sum_{n=2}^{\infty}\frac{n}{x^{n+1}})\,\mathrm {d} x \\\\ &= \int_0^{\infty} J(x) \frac{2x-1}{(x-1)^2x^2}\,\mathrm {d} x\end{align}

This is a quite interesting result which I believe is correct, but I'm still not satisfied. We can do better!