So this is a comment to my question. In general if $G/k$ is a connected linear algebraic $k$-group then one can show that it is generated by its Cartan subgroups defined over $k$.  (See the remark on the top of p. 220 of Borel's book on linear algebraic groups). For dimension reason we may take only finitely many of them. Finally, since I was assuming that $G$ is reductive it follows that every Cartan subgroup is a maximal torus and thus it is possible to generate $G$ with only finitely many tori.

So this is a comment to my question. In general if $G/k$ ($k$ is any field) is a connected linear algebraic $k$-group then one can show that it is generated over $\bar{k}$ by its Cartan subgroups defined over $k$.  (See the remark on the top of p. 220 of Borel's book on linear algebraic groups). For dimension reason we may take only finitely many of them. Finally, since I was assuming that $G$ is reductive it follows that every Cartan subgroup is a maximal torus and thus it is possible to generate $G$ with only finitely many $k$-tori.

Here is Brian Conrad's proof that was communicated to me through email in the case where $G$ is reductive and $k$ is infinite:

Consider a smooth connected affine $k$-group $G$. For every $X$ in $Lie(G)$ that is semisimple (relative to $G$, which is to say in the sense defined in Borel's book on algebraic groups), it is an important fact (proved in Borel's book on algebraic groups (see (**) in 9.1 and the Proposition just after it) that the scheme-theoretic centralizer $Z_G(X)$ of $X$ under the adjoint action of $G$ on $Lie(G)$ is smooth due to semisimplicity of $X$.

We proceed by induction on $dim(G)$, using the reductivity, and we work over infinite fields. (The case of finite fields is true, but requires other ideas.) We may assume $G$ isn't a torus, so it is not commutative and over $\bar{k}$ there are (by the classical structure theory over alg. closed fields) subgroups of type $SL_2$ or $PGL_2$ whose Lie algebras are not commutative (in any characteristic). Hence, $Lie(G)$ is not commutative. In the proof of 18.2(ii) in Borel's book he shows that $G$ is generated by finitely many of the centralizers $Z_G(X_i)^{\circ}$ for finitely many $X_i$ in $Lie(G)$ that are regular semisimple in a sense defined in 18.1. In particular, the discussion there implies (using that $Lie(G)$ is not commutative) that if $X\in Lie(G)$ is regular ss then $ad_G(X)$ is not zero and hence $Z_G(X)$ is not equal to $G$. Thus, the finitely many smooth connected $k$-subgroups $H_i = Z_G(X_i)^{\circ}$ that generate $G$ have strictly smaller dimension. But $Z_G(X)^{\circ}$ is always reductive for semisimple $X\in Lie(G)$ (13.19 in Borel's book), so by dimension induction we win. QED