Let $S = \{D\_1, D\_2, D\_3, \ldots \}$ be the set of all prime discriminants 
(or positive prime discriminants) of quadratic number fields. For such a 
discriminant let $\chi\_j(n) = (\frac{D\_j}n)$ be the associated Dirichlet 
character and 
$$ L(1,\chi\_j) = \sum\_{n \ge 1} \frac1n \Big(\frac{D\_j}{n}\Big) $$
the value of Dirichlet's L-series at $s = 1$. If we assume that
about half prime discriminants $D$ have $(D/p) = +1$ and the other half
satisfy $(D/p) = -1$, and if we interchange the limits, then the
geometric mean of the values of $L(1,\chi\_j)$ is given by 
\begin{align\*}
 \lim\_{k \to \infty} \bigg( \prod\_{j=1}^k L(1,\chi\_j) \bigg)^{1/k} 
 & = \lim\_k \prod\_p \Big(\frac{p}{p-1}\Big)^{\frac{k}{2k}}
 \cdot \Big(\frac{p}{p+1}\Big)^{\frac{k}{2k}} \\
 &= \prod\_p \Big(\frac{p^2}{p^2-1}\Big)^{1/2} = \sqrt{\zeta(2)} 
 = \frac{\pi}{\sqrt{6}}. 
\end{align\*}
Has this result due to Scholz been studied anywhere?


Scholz also believed that if $S$ denotes the set of all fundamental
discriminants, then the corresponding limit is equal to
$$ \prod \Big( \frac{p^2}{p^2-1} \Big)^{\frac{p}{2p+2}}. $$



Geometric Mean of $L(1,\chi)$ for quadratic Dirichlet characters
The distribution of values of $L(1,\chi\_d)$ as $d$ varies over fundamental discriminants has been extensively studied. For example, see this paper of [Granville and Soundararajan](http://arxiv.org/abs/math/0206031) which gives uniform such results (and discusses other references and history). The main result there shows that $L(1,\chi\_d)$ is distributed like a random Euler product $L(1,X)= \prod\_p (1-X(p)/p)^{-1}$ where $X(p)$ for primes $p$ denote independent random variables with $X(p)=1$ with probability $p/(2(p+1))$, $-1$ with probability $p/(2(p+1))$ and $0$ with probability $1/(p+1)$. From this the last assertion you make about fundamental discriminants follows: you want to compute (for large $D$)
$$ 
\exp\Big( \frac{1}{|\{|d|\le D\}|} \sum\_{|d|\le D} \log L(1,\chi\_d) \Big)
\sim \exp\Big({\Bbb E} (\log L(1,X)) \Big) = \prod\_p \Big(\frac{p^2}{p^2-1}\Big)^{\frac{p}{2p+2}}.
$$
Small modifications to the same techniques would allow you to study the family of prime discriminants that you mentioned -- the only difference is in adjusting the probabilistic model to reflect the fact that very few prime discriminants will be divisible a given prime $p$ (as opposed to all fundamental discriminants where this proportion is $1/(p+1)$). 


