Assuming AD, Woodin showed that no inner model that is missing a real correctly computes $\omega_1$. (This almost follows from Theorem 9 of Velickovic-Woodin's "Complexity of the set of reals of inner models of set theory.")
Therefore if you take an inner model $M$ of AD + $V = L(\mathbb R)$ whose $\omega_1$ is as small as possible, this model has no proper inner model of AD.
(This argument shows that inclusion is a wellfounded partial order of the inner models of $\text{AD}+ V = L(\mathbb R)$...)

~~If one generically wellorders the reals of $M$ without adding reals, one obtains a model of ZFC in which there is a minimum inner model of $\text{AD}$.~~ Update: this might be true, but as Dmytro Taranovsky points out in the comments, it is far from clear.

In general, however, there need not be a minimum inner model of $\text{AD}$. Assume there is a model of $\text{AD}$ containing only countably many reals. (This hypothesis is justified in the following paragraph.) Let $M$ be a minimal model of $\text{AD}$ with this property. Then there is a real $g$ that is $M$-generic for Cohen forcing. Let $N = L(\mathbb R)^{V[g]}$. By another theorem of Woodin, there is an elementary embedding $j : M\to N$. (There are more details in Kechris-Woodin's "Generic codes for uncountable ordinals.") Therefore $N$ is a minimal model of $\text{AD}$, since this is first order expressible, and $N$ is obviously distinct from $M$.

Of course, after forcing with $\text{Col}(\omega,\mathbb R)$ over a model of $\text{AD}^{L(\mathbb R)}$, there is an inner model of $\text{AD}$ that contains countably many reals, so the hypothesis of the above paragraph is consistent. In fact, the hypothesis is *true*: using $\mathbb R^\#$, one can build an inner model containing only countably many reals that is elementarily embeddable into $L(\mathbb R)$. (Take a countable elementary substructure of the mouse $\mathbb R^\#$ and then iterate away the top measure.) Therefore large cardinals imply that there is no minimum inner model of $\text{AD}$.