nt.number theory - Calculating the infinite product from the Hardy-Littlewood Conjecture F

I don't know what "the standard methods" means. Have you considered comparing
this with a more conventional Euler product? First, I will write $p$ instead of $varpi$ since
I don't like $varpi$ (looks too much like $overline{omega}$ for my tastes).

Expanding the $p$-th factor as a geometric series,
$$
1-frac{1}{p-1}left(frac Dpright) =
1-frac{1/p}{1-1/p}left(frac Dpright) =
1-frac{1}{p}left(frac Dpright) - frac{1}{p^2}left(frac Dpright) - cdots,
$$
and the first two terms have a product that is known:
$prod_{p} (1 - (D|p)/p) = 1/L(1,chi_D)$, where $chi_D(p) = (frac{D}{p})$ for odd $p$.
(I'm being a bit sloppy about the Euler factor at 2.) Therefore if we divide the $p$-th factor in the Hardy--Littlewood constant by $1 - (D|p)/p$ then we'll have the $p$-th term be $1 + O(1/p^2)$, which has better convergence (not ultra-fast, but better than what you had before.) We write
$$
prod_{p} left(1 - frac{(D|p)}{p-1}right) =
prod_{p} frac{1 - (D|p)/(p-1)}{1 - (D|p)/p} cdot
left(1 - frac{(D|p)}{p}right)=
prod_{p} frac{1 - (D|p)/(p-1)}{1 - (D|p)/p} cdot prod_{p}
left(1 - frac{(D|p)}{p}right).
$$
This last expression is
$$
frac{1}{L(1,chi_D)}prod_{p} frac{1 - (D|p)/(p-1)}{1 - (D|p)/p} =
frac{1}{L(1,chi_D)}prod_{p} left(1 + Oleft(frac{1}{p^2}right)right).
$$

There are ways of accelerating these quasi-Euler products further. See
Pieter Moree, Approximation of singular series and automata,
Manuscripta Math. 101 (2000), 385--399.

Last comment: yes, when $chi$ is nontrivial the Euler product of $L(s,chi)$ at $s = 1$ does equal the $L$-function at $s = 1$. That requires an argument, since $s=1$ is not in the range of absolute convergence of the Euler product.