If $T$ is taken to be a maximal torus of $G$ lying in $B$, the question may be reformulated using the set-up in 1.1-1.2 of the fundamental paper: Representations of Reductive Groups Over Finite Fields, P. Deligne and G. Lusztig, The Annals of Mathematics, Second Series, Vol. 103, No. 1 (Jan., 1976), pp. 103-161 (available in JSTOR). Relative to any such fixed choice of the pair $(T,B)$, they identify the $G$-orbits in $G/B \times G/B$ with the elements of a canonical Weyl group (independent of choices). On the other hand, an old result of Chevalley identifies the $T$-fixed points of $G/B$ with the set of Borels containing $T$, thus with the Weyl group $N(T)/T$ of $G$ relative to $T$. So you are starting with a pair of such elements, which in the Deligne-Lusztig set-up determine a single element of the "absolute" Weyl group. I'm not immediately sure how your question about the closure of a $B$-orbit will translate into this framework, but looking at it this way may be helpful.

However, if $T$ is arbitrary, the question seems to become much harder.

If $T$ is taken to be a maximal torus of $G$ lying in $B$, the question may be reformulated using the set-up in 1.1-1.2 of the fundamental paper: Representations of Reductive Groups Over Finite Fields, P. Deligne and G. Lusztig, The Annals of Mathematics, Second Series, Vol. 103, No. 1 (Jan., 1976), pp. 103-161 (available in JSTOR). Relative to any such fixed choice of the pair $(T,B)$, they identify the $G$-orbits in $G/B \times G/B$ with the elements of a canonical Weyl group (independent of choices). On the other hand, an old result of Chevalley identifies the $T$-fixed points of $G/B$ with the set of Borels containing $T$, thus with the Weyl group $N(T)/T$ of $G$ relative to $T$. So you are starting with a pair of such elements, which in the Deligne-Lusztig set-up determine a single element of the "absolute" Weyl group. I'm not immediately sure how your question about the closure of a $B$-orbit will translate into this framework, but looking at it this way may be helpful. (Special cases suggest that you may just get a copy of a Schubert variety, but this is probably oversimplified.)