Consider a finite Galois extension $L$ of $\mathbb Q$, of Galois group $G$. Let $k \geq 1$ be a fixed integer. Let $D$ be a subset of $G^k$ invariant by conjugation and by the natural action of the symmetric group $S_k$ on $G^k$.

Let $A=A(L,D)$ be the set of all integers $n$ which are products of $k$ primes $n=p_1\dots p_k$ (variant: $k$ distinct primes) such that $(Frob_{p_1},\dots,Frob_{p_k}) \in D.$

(Here $Frob_p$ is a Frobenius at $p$ in $G$, assuming $p$ is unramified in $L$. if one of the $p_i$ is ramified, by convention say that $n$ is not in $A$.)

Let $A(x)$ be the number of integers $n<x$ in $A$.

Is there a known equivalent for $A(x)$ when $x$ goes to infinity?

Note that the case $k=1$ is Chebotarev's density theorem. In this case, $A(x) \sim \frac{|D|}{|G|} x/\log x $. In the general case, I would expect $$A(x) \sim c x/\log x \frac{(\log \log x)^{k-1}}{(k-1)!},$$ with a constant $c$ between $0$ and $1$ depending of $D$, $G$ (and perhaps only of $|D|$, $|G|$), based on the case $k=1$ and the case $D=G^k$, where $A$ is the set of all $k$-almost primes and the desired equivalent is a classical theorem of Landau. Because of these illustrious limit cases, I would not be surprised if this question was dealt with somewhere in the literature. But I couldn't find where.

I would already be interested in the case where $G$ is abelian. In this case the question takes a more classical form, which I explicit in case some readers are uncomfortable with Chebotarev. We can assume that $G=(\mathbb Z/a\mathbb Z)^\ast$. Then $D$ is a subset of $G^k$ invariant by permutation of the coordinates. The set $A$ is then the set of integers $n=p_1\dots p_k$ such that $(p_1 \pmod{a},\dots,p_k \pmod{a})$ is in $D$, and the question is still to find an equivalent for the number $A(x)$ of such integers $n$ which are $\leq x$.

**Edit**: I have made some computations in the simplest non trivial abelian case. This is the case $k=2$, and $L=Q(i)$ so that $G={1,-1}$ and $Frob_p$ for an odd $p$ is just $p \mod{4}$.
You have $3$ basic subsets $D$ of $G^2$ invariant by $S_2$ which are
$D_1 = \{(1,1)\}$, $D_0=\{(1,-1),(-1,1)\}$, $D_{-1}=\{(-1,-1)\}$.
The corresponding sets of integers are the sets $$A_1=\{2-\text{almost primes } n=pq,\ \ p\equiv q \equiv 1 \pmod{4}\}$$ $$A_0 =\{2-\text{almost primes } n,\ \ n\equiv -1 \pmod{4}\}$$ $$A_{-1}=\{2-\text{almost primes } n=pq,\ \ p\equiv q \equiv -1 \pmod{4} \}.$$ I strongly expect in this case that the $A_{-1},A_1,A_0$ take respectively a proportion $1/4,1/4,1/2$ (or $|D|/|G|^2$) of all $2$-almost primes. If true, this is probably not to hard to prove by the method indicated by Lucia in comment. However, I have not been able to "see this" in the computation. For $x=10^7$, for instance, those proportions are $0.300,0.200,0.500$; for $x=5 \cdot 10^7$, they are $0.292,0.500,0.208$. I think the domination of products of two primes congruent to $-1$ above the products of two primes congruent to $1$ is well explained by the "primes race": the advantage of primes congruent to $-1$ among small primes is felt even more for the product of two primes. But this has discouraged me of making more computations to guess the correct value of the constant $c$...