Consider a supplement $T'$ to $T$, so that $T \times T'$ is a maximal torus of $G$. The centraliser $H_0$ of $T'$ in $G$ is then a reductive group, defined over $Q$ if $T'$ is, whose maximal subgroup is $T \times T'$. Taking $H$ the derived group of $H_0$ will kill the central $T'$ and leave you with a group having $T$ as maximal torus — it is not centralised as you can check by constructing an appropriate $\mathfrak{sl}_2$-triple.

Consider a supplement $T'$ to $T$, so that $T \times T'$ is a maximal torus of $G$. The centraliser $H_0$ of $T'$ in $G$ is then a reductive group, defined over $Q$ if $T'$ is, whose maximal subgroup is $T \times T'$. Taking $H$ the derived group of $H_0$ will kill the center of $H_0$. $T' \times T$ meets $H$ against a maximal torus of $H$, which is the intersection of the roots of $G$ whose kernel contains $T$.

Using this process, you do not really get $T$ as maximal torus, but rather a kind of “arithmetic closure” of it. Is this close enough?

Consider a supplement $T'$ to $T$, so that $T \times T'$ is a maximal torus of $G$. The centraliser $H$ of $T'$ in $G$ is then a reductive group, defined over $Q$ if $T'$ is, and $T$ must not be in the centre of $H$ — and will actually never be as you can check by constructing an appropriate $\mathfrak{sl}_2$-triple.

Consider a supplement $T'$ to $T$, so that $T \times T'$ is a maximal torus of $G$. The centraliser $H_0$ of $T'$ in $G$ is then a reductive group, defined over $Q$ if $T'$ is, whose maximal subgroup is $T \times T'$. Taking $H$ the derived group of $H_0$ will kill the central $T'$ and leave you with a group having $T$ as maximal torus — it is not centralised as you can check by constructing an appropriate $\mathfrak{sl}_2$-triple.