Kevin Buzzard informs me, by e-mail and in the above comment, that what he cares about is the case where $A$ is regular. In particular, it would be good to know whether every $T$ is a matrix ring in this case. I thought about it but didn't solve this, so here is a record of my ideas. Let $n$ be the dimension of $A$.

(1) A subring of a matrix is obviously a matrix ring. So, if we knew that $T$ was a matrix ring whenever $T$ was normal, this would establish that $T$ was always a matrix ring.

(2) As explained in my previous answer, if $T$ is free as an $A$-module, then $T$ is a matrix ring. We have the following implications: if $T$ is Cohen-Macaulay then $T$ torsion-free and finite over $A$ implies $T$ flat over $A$; if $A$ is a polynomial ring then $T$ flat over $A$ implies $T$ free over $A$. So, if $A$ is a polynomial ring and $T$ is Cohen-Macaulay, then $T$ is a matrix ring.

In particular, if $n=1$ or $2$, then $T$ normal implies $T$ Cohen-Macaulay. So, in these cases, and with $A$ a polynomial ring, $T$ is a matrix ring.

(3) As explained above, if $T^\*$ is free over $A$, we also get to conclude that $T$ is a matrix ring. Unfortunately, this can fail when $n \geq 3$.

(4) If there is some $T$-module $M$ on which $T$ acts without kernel, and $T$ is free as an $A$-module, then $T$ is a matrix algebra. Restated geometrically, if there is any coherent sheaf on $\mathrm{Spec}(T)$, with support on the whole of $\mathrm{Spec}(T)$, whose pushforward to $\mathrm{Spec}(A)$ is a trivial vector bundle, then $T$ is a matrix algebra. If we restrict our attention to $A$ a local ring, or a polynomial ring, then the adjective "trivial" comes for free.

So, if there were to be a counter-example, we would want $n \geq 3$ and we would want $T$ to be normal but not Cohen-Macaulay. Moreover, we would want that basically any $T$-module is not free as an $A$-module.

Any ideas?