Here's an argument that if $A$ is a finite set of algebraic numbers, then $d_A(n)$ grows at most exponentially.

Suppose $A$ is contained in some number field $K$. For $x\in K^\times$ there's a product formula
$$
\prod_v |x|_v=1,
$$
where $v$ runs over the valuations on $K$ and $|\cdot|_v$ is a normalized absolute value. In order 
for some absolute value to be small, the product of the others must be large.

For fixed $A$, there's a finite set of places $S$ of $K$ such that $|x|_v\leq 1$ for all $x\in S_A(n)$ whenever $v\not\in S$ ($S$ consists of all the Achimedean places, and the finite places dividing denominators of elements of $A$). It's not hard to see that for fixed $v$, $\max \{|x|_v:x\in S_A(n)\}$ grows at most exponentially in $n$. Since the number of $v$ for which $|x|_v>1$ is bounded, we get lower bound of exponential decay the minimal absolute value of elements of $S_A(n)$.