$\newcommand\R{\mathbb R}\newcommand\la{\lambda}$A measure $\mu$ is Ahlfors regular (according to your definition) iff it has a density (with respect to the Lebesgue measure $\la$) bounded away from $0$ and $\infty$ (if $B(x,r)$ denotes the ball centered at $x$ and of radius $r$).

Indeed, by inscribing and circumscribing, we can replace $B(x,r)$ in your definition by the (left-open) $n$-cube $b(x,r):=\prod_1^n(x_i,x_i+r]$ (say of size $r$) for $x=(x_1,\dots,x_n)\in\R^n$ (at the same time also appropriately changing the values of $c$ and $C$), so that $$ c r^d\le\mu(b(x,r))\le Cr^d \tag{1}\label{1} $$ for all $x\in\R^n$ and $r>0$.

Partitioning, in the natural manner, the unit $n$-cube $b(0,1)$ into $N^n$ $\ n$-cubes each of size $1/N$, from \eqref{1} we get $$ c N^{n-d}\le\mu(b(0,1))\le CN^{n-d} $$ for all $N$, which is possible only when $d=n$.

Therefore and by approximation, we get $$ c \prod_1^n r_i\le\mu\Big(\prod_1^n(x_i,x_i+r_i]\Big)\le C \prod_1^n r_i $$ for any real $x_1,\dots,x_n$ and any positive real $r_1,\dots,r_n$. That is, $c\la\le\mu\le C\la$ on the semiring of all left-open $n$-boxes.

We conclude that $c\la\le\mu\le C\la$ on the Lebesgue $\sigma$-algebra, as claimed.

$\newcommand\R{\mathbb R}\newcommand\la{\lambda}$A measure $\mu$ is Ahlfors regular (according to your definition) iff it has a density (with respect to the Lebesgue measure $\la$) bounded away from $0$ and $\infty$ (if $B(x,r)$ denotes the ball centered at $x$ and of radius $r$).

Indeed, by inscribing and circumscribing, we can replace $B(x,r)$ in your definition by the (left-open) $n$-cube $b(x,r):=\prod_1^n(x_i,x_i+r]$ (say of size $r$) for $x=(x_1,\dots,x_n)\in\R^n$ (at the same time also appropriately changing the values of $c$ and $C$), so that $$ c r^d\le\mu(b(x,r))\le Cr^d \tag{1}\label{1} $$ for all $x\in\R^n$ and $r>0$.

Partitioning, in the natural manner, the unit $n$-cube $b(0,1)$ into $N^n$ $\ n$-cubes each of size $1/N$, from \eqref{1} we get $$ c N^{n-d}\le\mu(b(0,1))\le CN^{n-d} \tag{2}\label{2} $$ for all $N$, which is possible only when $d=n$ -- otherwise, $N^{n-d}$ will go either to $0$ or to $\infty$ as $N\to\infty$, which would contradict one of the two inequalities in \eqref{2}, since $\mu(b(0,1))\in(0,\infty)$.

Therefore and by partitioning and approximation, we get $$ c \prod_1^n r_i\le\mu\Big(\prod_1^n(x_i,x_i+r_i]\Big)\le C \prod_1^n r_i $$ for any real $x_1,\dots,x_n$ and any positive real $r_1,\dots,r_n$. That is, $c\la\le\mu\le C\la$ on the semiring of all left-open $n$-boxes.

We conclude that $c\la\le\mu\le C\la$ on the Lebesgue $\sigma$-algebra, as claimed.