As it seems to be of some interest, let me record some comments on the convergence rate of the Euler product (which are probably standard).

We consider $$C(f)=\prod_p \frac{1-\eta_f(p)/p}{1-1/p},$$ and as was pointed out, this is kind of like an $L$-function at $s=1$, namely that at good primes the Euler factor $(1-\eta_f(p)/p^s)^{-1}$ gives the same Dirichlet series coefficient at $p$ (though not powers of $p$) as the Dedekind $\zeta$-function of the field associated to $f$.

Everything below will be in terms of relative precision, which should not matter much as $C(f)$ itself is not particularly big nor small. Consider the contribution in the above product from $p\ge X$ and take logs, getting $$\sum_{p\ge X}\log(1-\eta_f(p)/p)-\log(1-1/p)=\sum_{p^k\ge X}\frac{1-\eta_f(p)^k}{p^k}=\sum_{p\ge X}\frac{1-\eta_f(p)}{p}+O(1/X),$$ where the big-Oh constant should be no worse than about the square of the degree of $f$. Under GRH and the Artin conjecture, we will show this $p$-sum tail is $O((\log |\Delta_f|)/\sqrt X)$.

As indicated above, $\eta_f(p)$ relates to Dedekind $\zeta$-function for $f$. Let us write $A(s)=\zeta_f(s)/\zeta(s)$, and this Artin $L$-function will be "nice" if we assume enough hypotheses. In particular, by Perron's formula we have $$\int_{(2)}X^{s-1}\frac{A'}{A}(s){ds\over 2\pi is}= -\sum_{\|{\frak p}\|^k\le X}{\log\|{\frak p}\|\over \|{\frak p}\|^k} +\sum_{p^k\le X}{\log p\over p^k}.$$ The terms with $k\ge 2$ contribute $B_1+B_2+O(1/X)$, where $B_i$ are the relevant sums extended to infinity (not cut off at $X$), and the prime ideals with $\|{\frak p}\|\neq p$ contribute $O(1/\sqrt X)$.

The prime ideals with $\|{\frak p}\|=p$ are those counted by $\eta_f(p)$ (at least for good primes), so we have $$\int_{(2)}X^{s-1}\frac{A'}{A}(s){ds\over 2\pi is}=B_1+B_2+\sum_{p\le X}(1-\eta_f(p))\frac{\log p}{p}+O(1/\sqrt X).$$ We then truncate the line of integration at say height $X^3$, with error $O(X/X^3)$ (again the big-Oh depends on the degree of $f$). We then move to contour to the left, say $\sigma=1/4$. Under GRH the zeros are on $\sigma=1/2$, and the contribution from $\sigma=1/4$ is $O((\log N)(\log X)/X^{3/4})$, where $N$ is the conductor, thus the absolute value of the discriminant of $f$. Similarly the zeros themselves contribute $O(\log N)(\log X)/\sqrt X$, by zero-density estimates (namely $\log N$) and a harmonic sum up to height $X^3$ from the $1/s$.

So we get $$\sum_{p\le X}(1-\eta_f(p))\frac{\log p}{p}=B+O((\log N)(\log X)/\sqrt X)$$ and by partial summation $$\sum_{p\le X}\frac{1-\eta_f(p)}{p}=\tilde B+O((\log N)/\sqrt X).$$ In particular we have $$\sum_{p\ge X}\frac{1-\eta_f(p)}{p}=O((\log N)/\sqrt X),$$ which as above gives the convergence rate for the desired constant $C(f)$.

Maybe some analytic number theorist will come around and correct anything (and fix any sign errors I made).

As it seems to be of some interest, let me record some comments on the convergence rate of the Euler product (which are probably standard).

We consider $$C(f)=\prod_p \frac{1-\eta_f(p)/p}{1-1/p},$$ and as was pointed out, this is kind of like an $L$-function at $s=1$, namely that at good primes the Euler factor $(1-\eta_f(p)/p^s)^{-1}$ gives the same Dirichlet series coefficient at $p$ (though not powers of $p$) as the Dedekind $\zeta$-function of the field associated to $f$.

Everything below will be in terms of relative precision, which should not matter much as $C(f)$ itself is not particularly big nor small. Consider the contribution in the above product from $p\ge X$ and take logs, getting $$\sum_{p\ge X}\log(1-\eta_f(p)/p)-\log(1-1/p)=\sum_{p^k\ge X}\frac{1-\eta_f(p)^k}{p^k}=\sum_{p\ge X}\frac{1-\eta_f(p)}{p}+O(1/X),$$ where the big-Oh constant should be no worse than about the square of the degree of $f$. Under GRH and the Artin conjecture, we will show this $p$-sum tail is $O((\log |\Delta_f|)/\sqrt X)$.

As indicated above, $\eta_f(p)$ relates to Dedekind $\zeta$-function for $f$. Let us write $A(s)=\zeta_f(s)/\zeta(s)$, and this Artin $L$-function will be "nice" if we assume enough hypotheses. In particular, by Perron's formula we have $$\int_{(1)}X^s\frac{A'}{A}(s+1){ds\over 2\pi is}= -\sum_{\|{\frak p}\|^k\le X}{\log\|{\frak p}\|\over \|{\frak p}\|^k} +\sum_{p^k\le X}{\log p\over p^k}.$$ The terms with $k\ge 2$ contribute $B_1+B_2+O(1/X)$, where $B_i$ are the relevant sums extended to infinity (not cut off at $X$), and the prime ideals with $\|{\frak p}\|\neq p$ contribute $O(1/\sqrt X)$.

The prime ideals with $\|{\frak p}\|=p$ are those counted by $\eta_f(p)$ (at least for good primes), so we have $$\int_{(1)}X^s\frac{A'}{A}(s+1){ds\over 2\pi is}=B_1+B_2+\sum_{p\le X}(1-\eta_f(p))\frac{\log p}{p}+O(1/\sqrt X).$$ We then truncate the line of integration at say height $X^3$, with error $O(X/X^3)$ (again the big-Oh depends on the degree of $f$). We then move to contour to the left, say $\sigma=-3/4$. Under GRH the zeros are on $\sigma=-1/2$, and the contribution from $\sigma=-3/4$ is $O((\log N)(\log X)/X^{3/4})$, where $N$ is the conductor, thus the absolute value of the discriminant of $f$. Similarly the zeros themselves contribute $O(\log N)(\log X)/\sqrt X$, by zero-density estimates (namely $\log N$) and a harmonic sum up to height $X^3$ from the $1/s$.

So we get $$\sum_{p\le X}(1-\eta_f(p))\frac{\log p}{p}=B+O((\log N)(\log X)/\sqrt X)$$ and by partial summation $$\sum_{p\le X}\frac{1-\eta_f(p)}{p}=\tilde B+O((\log N)/\sqrt X).$$ In particular we have $$\sum_{p\ge X}\frac{1-\eta_f(p)}{p}=O((\log N)/\sqrt X),$$ which as above gives the convergence rate for the desired constant $C(f)$.

Maybe some analytic number theorist will come around and correct anything (and fix any sign errors I made). Also, the Artin conjecture might not be necessary, just the GRH for $\zeta_f$ and $\zeta$ (poles on the half-line are not bothersome).