I'm assuming a Li-Yorke pair for $T$ is a pair $x,y$ such that $\limsup_{n\to \infty} d(T^n(x), T^n(y)) > 0 $ and $\liminf_{n\to \infty} d(T^n(x), T^n(y)) = 0$.
For toral automorphisms, I think the answer to your question is no, based on this:
Claim: A linear toral automorphism has a Li-Yorke pair if and only if it is ergodic.
Since ergodic linear toral automorphisms have positive entropy (see for instance here), the claim answers your question negatively.
Proof of the claim: The if part is easy, since a point with a dense orbit and a fixed point give a Li-Yorke pair. Now assume $T$ is not ergodic but has a Li-Yorke pair. Being non-ergodic implies that there is an eigenvalue which is a root of the unity. We may replace $T$ by some power of $T$ and assume that $1$ is an eigenvalue. This implies that the dynamics fibers over a circle: you can find an integer eigenvector with eigenvalue $1$, and after a linear change of coordinates you may assume the dynamics is $(x,y) \mapsto (x, T'y)$ for $(x,y)\in \mathbb{T}\times \mathbb{T}^{d-1}$. Clearly if you have a Li-Yorke pair, both points must belong to the same fiber, and the automorphism $T'$ of $\mathbb{T}^{d-1}$ also has a Li-Yorke pair. So you have reduced the dimension. The claim is trivial in dimension $1$, so the proof is completed by induction.