What can be said about $\prod_{s=2}^{\infty} \zeta(s) $?

Marco's answer is about how to compute $v$ with desired precision.

My answer is to provide the meaning of the number. Let $A_n$ be the number of nonisomorphic abelian groups of order $n$. A result by Erdos and Szekeres shows that

$$ \sum_{n\leq x} A_n = v x + O(\sqrt x), $$ where $v=\prod_{k=2}^{\infty} \zeta(k)$. Therefore, we can say that $v$ is the mean value of the number of nonisomorphic abelian groups of order $n$.

The proof can be done in two parts. The steps are outlined in Montgomery and Vaughan's Multiplicative Number Theory I. (2.1 #19).

(a) Show that $\sum_{n=1}^{\infty} A_n n^{-s}=\prod_{k=1}^{\infty} \zeta(ks)$.

(b) Show that $\sum_{n\leq x} A_n = vx + O(\sqrt x)$ where $v=\prod_{k=2}^{\infty} \zeta(k)$.

Proof of (a).

For any $n$, consider prime factorization of $n$. For each $p^e||n$, consider a partition $e=\lambda_1+\lambda_2+\cdots+\lambda_r$ with $\lambda_i\leq \lambda_j$ for $i\leq j$. Let this correspond to the $p$-primary part $\mathbb{Z}/p^{\lambda_1}\mathbb{Z}\times\mathbb{Z}/p^{\lambda_2}\mathbb{Z}\times\cdots\times\mathbb{Z}/p^{\lambda_r}\mathbb{Z}$ of an abelian group of order $n$.

Thus, $\sum_{n=1}^{\infty}A_n n^{-s}$ should be $$\begin{align} \prod_p \prod_{k=1}^{\infty} \left( 1- \frac1{p^{ks}}\right)^{-1}&=\prod_p\left(1-\frac1{p^s}\right)^{-1}\left(1-\frac1{p^{2s}}\right)^{-1}\cdots \left(1-\frac1{p^{ks}}\right)^{-1} \cdots \\ &=\prod_{k=1}^{\infty} \zeta(ks), \ \ \Re(s)>1.\end{align} $$

Proof of (b).

By (a), we have $\sum_{n=1}^{\infty} A_n n^{-s} = \zeta(s) \zeta(2s) F(s)$, where $F(s) = \prod_{k=3}^{\infty} \zeta(ks)= \sum_{n=1}^{\infty} f(n)n^{-s}$, $\Re(s)>1/3$. Then $$ \begin{align} \sum_{n\leq x} A_n &= \sum_{mdk\leq x} \delta_2(d) f(m)\\ &=\sum_{m\leq x} f(m) \sum_{d\leq \frac x{m}}\delta_2(d)\sum_{k\leq \frac x{md}} 1\\ &=\sum_{m\leq x} f(m)\sum_{d\leq \frac x{m}}\delta_2(d)\left(\frac{ x}{md}+O(1)\right)\\ &=\sum_{m\leq x}f(m)\frac xm \left(\zeta(2)+O(\sqrt{\frac mx})\right)+O(\sqrt x)\\ &=xF(1)\zeta(2)+O(\sqrt x). \end{align}$$ Here, $\delta_2(d)=1$ if $d$ is a square, and $0$ otherwise. The result follows by $v=F(1)\zeta(2)$.

You can compute this product with arbitrary precision. From the well known inequality $$\zeta\left(s\right)-\sum_{n=1}^{N}\frac{1}{n^{s}}<\frac{N^{1-s}}{s-1},\,s>1$$ we have $$\sum_{n>N}\log\left(\zeta\left(n\right)\right)<\sum_{n>N}\left(\zeta\left(n\right)-1\right)<\sum_{n>N}\frac{1}{2^{n}}\left(1+\frac{2}{n-1}\right)<2^{-N}\left(1+\frac{2}{N}\right).$$