The Laurent polynomial ring is fairly trivial. If $n=1$, clearly $A=k$ and if $n=2$, $A$ is rational and smooth, so $A=k[u, f(u)^{-1}]$ and the rest should be clear. So, let us assume $n\geq 3$.
Under the isomorphism, $T=\alpha X_1^{a_1}\cdots X_n^{a_n}$ for some $\alpha\in k^*$ and $a_i\in\mathbb{Z}$. One first checks that the gcd of the $a_i$'s is one. If not, $T=\alpha z^d, z\in A[T,T^{-1}]^*$, $d>1$. But, the unit group of $A[T,T^{-1}]$ is $A^*\times\mathbb{Z} T$. Easy to see that $d$ has to be one.
Since $n\geq 3$, we can find integers $b_i, 2\leq i\leq n$ so that $a_1+\sum_{i=2}^n a_ib_i$=1. Consider the automorphism $k[X_i, X_i^{-1}1\leq i\leq n]$ given by $X_1\mapsto X_1$, $X_i\mapsto X_iX_1^{b_i}$ for $i>1$. Composing, we see that $T\mapsto \alpha X_1^{a_1}X_2^{b_2}X_1^{a_2b_2}\cdots X_n^{b_n}X_1^{a_nb_n}= \alpha X_1X_2^{b_2}\cdots X_n^{b_n}$. Now, another easy automorphism will arrange it so that $T\mapsto X_1$. Then it is clear that $A\cong k[X_i,X_i^{-1}, 1<i\leq n]$.