My original answer was flawed. I am replacing it with the following.

The irreducible representations $M_\lambda$ of the symmetric group
$S_n$ are indexed by partitions $\lambda$ of $n$. If $\lambda\neq
\lambda'$ (the conjugate partition to $\lambda$), then the restriction
$N_\lambda$ of $M_\lambda$ to $A_n$ remains irreducible. If
$\lambda=\lambda'$ then $M_\lambda$ splits into two irreps of the same
dimension.

Now $n!/\dim(M_\lambda)$ is the product $H_\lambda$ of the hook
lengths of $\lambda$. Thus $n!/(\dim N_\lambda)$ is either the product
of the hook lengths (when $\lambda\neq \lambda'$) or twice this
product. We are interested in those $\lambda$ that minimize $n!/(\dim
N_\lambda)$. Suppose that $\lambda$ minimizes $n!/(\dim N_\lambda)$
and $\lambda=\lambda'$. Thus this minimum value is
$2H_\lambda$. Remove any corner square from $\lambda$, giving a
partition $\mu$ of $n-1$. Then clearly $H_\mu<2H_\mu<2H_\lambda$. Now
suppose that $\lambda\neq\lambda'$. If $\mu=\mu'$ then we need to show
$2H_\mu<H_\lambda$, which is not always true. However, it is easy to
see that we can in fact remove a corner square from $\lambda$ so that
$\mu\neq \mu'$ ($n\geq 3$), so $(n-1)!/(\dim N_\mu)=
H_\mu<H_\lambda=n!/(\dim N_\lambda)$. Thus it seems to me that the
numbers $(n!/2)/\mathrm{cd}_n$ are strictly increasing for $n\geq 3$.