Take any $x\in X$ with $m:=\mu(\{x\})\in(0,\infty)$ (since $X$ is finite and $\mu$ is a probability measure, such a point $x$ exists). Let $R:=\min\{d(y,x)\colon y\in X\setminus\{x\}\}$. Then $R\in(0,\infty)$ and $B(x,r)=\{x\}$ for $r\in(0,R)$.

Letting now $r\downarrow0$, from $cr^q \le \mu(B(x,r)) \le Cr^q$ we get $m\le 0$ if $q>0$ and $\infty\le m$ if $q<0$. So, we get a contradiction with $m\in(0,\infty)$ unless $q=0$.

So, $\mu$ can be Ahlfors $q$-regular only if $q=0$.

Take any $x\in X$ with $m:=\mu(\{x\})\in(0,\infty)$ (since $X$ is finite and $\mu$ is a probability measure, such a point $x$ exists). Let $R:=\min\{d(y,x)\colon y\in X\setminus\{x\}\}$. Then $R\in(0,\infty)$ and $B(x,r)=\{x\}$ for $r\in(0,R)$.

Letting now $r\downarrow0$, from $cr^q \le \mu(B(x,r)) \le Cr^q$ we get $m\le 0$ if $q>0$ and $\infty\le m$ if $q<0$. So, we get a contradiction with $m\in(0,\infty)$ unless $q=0$.

So, $\mu$ can be Ahlfors $q$-regular only if $q=0$.

On the other hand, any probability measure over a finite set is clearly Ahlfors $0$-regular (assuming the convention that $r^q=1$ if $r=0=q$).

Thus, any probability measure over a finite set is Ahlfors $q$-regular iff $q=0$.