The situation is similar to the one of closed manifolds. One defines "boundary-prime" manifolds as those that cannot be decomposed nontrivially in a boundary-connected sum.
Note that if $M$ is connected, has nonempty boundary and is not prime, then $M$ is never boundary-prime. Namely, take a 2-sphere $S\subset M$ separating $M$ into two components none of which is a ball. Connect $S$ to $\partial M$ by a 1-handle $D^2\times [0,1]$ ($D^2\times \{0\}\subset S$, $D^2\times \{1\}\subset \partial M$). Now, remove from $S$ the open disk equal to $int(D^2)\times \{0\}$ and add to $S$ the annulus $\partial D^2\times [0,1]$. The resulting surface $S'$ is a 2-dimensional disk with $\partial S'\subset \partial M$. This disk will cut $M$ in two components, none of which is a ball.
This works no matter what $\partial M$ is, in particular, if $\partial M=T^2$.
Thus, in what follows (until the concluding paragraph), I will assume that $M$ is prime.
Lemma. If $\partial M$ is a torus, then $M$ is necessarily boundary-prime.
Proof. Let $D\subset M$ be a properly embedded disk splitting $M$ in two components, none of which is a 3-ball. (Splitting means that you remove from $M$ an open tubular neighborhood of $D$.) Since $D$ separates $M$, $\partial D$ separates $T^2=\partial M$, hence, bounds a disk $D'\subset T^2$. Taking the union $D\cup D'$ we obtain a non-properly embedded 2-sphere in $M$. Pushing the disk $D'$ slightly into $M$, we obtain a 2-sphere $S\subset M$ disjoint from the boundary. Since $D$ was splitting $M$ into two submanifolds none of which is a 3-ball, the same holds for $S$. Hence, $M$ is not prime, contradicting the standing assumption. qed
We continue the discussion of prime manifolds $M$ with toral boundary. Such a manifold still can have compressible boundary. However, if this is the case, a boundary-compressing $D$ disk in $M$ is necessarily nonseparating. (Since every separating loop in the torus $T^2$ bounds a disk in $T^2$.) Cutting $M$ open along $D$ results in a manifold $M'$ with spherical boundary. If $M'$ is homeomorphic to the 3-ball, then $M$ itself is a solid torus, $\hat{T}=D^2\times S^1$. Otherwise, attaching $B^3$ along $\partial M'$ results in a closed 3-manifold $N$ which is not $S^3$. Then $M= N\# \hat{T}$. But there is one more possibility, namely, $\partial M$ is incompressible. There are many manifolds like that, for instance, the exterior of any nontrivial knot in $S^3$.
Hence, we obtain the trichotomy for 3-manifolds $M$ which need not be $\partial$-prime.
To conclude: Suppose that $M$ is a connected (not necessarily prime) 3-manifold and $\partial M$ is homeomorphic to $T^2$. Then one of the following mutually exclusive properties holds:
$M$ is not prime, equivalently, is not $\partial$-prime.
$M=\hat{T}$.
$M$ is prime and $\partial M$ is incompressible.