I'm hoping someone will check this computation. I get that the Bott-Samelson over $SL_3/B$ is not homeomorphic to $(\mathbb{P}^1)^3$. That is a major obstacle for the approach that Allen and VA discuss above.

There are two reduced words for the longest element of $S_3$, but the automorphism of the Dynkin diagram exchanges them. So, as an abstract variety, it makes sense to talk about the Bott-Samelson for $SL_3$. We'll call it $X$.

Let $F$ be $\mathbb{P}^2$ blown up at a point, with $\pi: F \to \mathbb{P}^2$ the blowdown. Then $X$ is a $\mathbb{P}^1$ bundle over $F$. Specifically, let $Q$ be the tautological quotient over $\mathbb{P}^2$. I believe that $X$ is $\mathbb{P}(\pi^* Q)$.

Let's compute $H^*(X)$. Everything is in even degree. $H^*(\mathbb{P}^2) = \mathbb{Z}[H]/H^3$, where $H$ is the hyperplane class. The blowup $F$ has $H^*(F) = \mathbb{Z}[H,E]/\langle H^2+E^2,\ HE \rangle$ where $E$ is the class of the exceptional fiber. The chern class of $Q$ is $1-H+H^2$. So $$H^(X) = H^*(F)[Z]/\langle Z^2 - ZH + H^2 \rangle = \mathbb{Z}[H,E,Z]/\langle H^2+E^2,\ HE,\ Z^2-ZH+H^2 \rangle.$$

So $H^2(X)$ is three dimensional. We have a cubic form on $H^2(X)$ given by $\alpha \mapsto \int_X \alpha^3$. I get that this cubic form is $$\int_X \begin{pmatrix} Z^3 & & & \\ Z^2 H & Z^2 E & & \\ Z H^2 & ZHE & Z E^2 & \\ H^3 & H^2 E & H E^2 & E^3 \end{pmatrix}= \begin{pmatrix} 0 & & & \\ 1 & 0 & & \\ 1 & 0 & -1 & \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Of course, the corresponding matrix for $(\mathbb{P}^1)^3$ is $$\begin{pmatrix} 0 & & & \\ 0 & 0 & & \\ 0 & 1 & 0 & \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Now, if there were a homeomorphism $X \cong (\mathbb{P}^1)^3$, then there would be an isomorphism $ H^2(X) \otimes \mathbb{C} \cong H^2((\mathbb{P}^1)^3) \otimes \mathbb{C}$ taking one cubic to the other. But I get that the first cubic is a line times a conic, while the second is three lines.

I'm hoping someone will check this computation. I get that the Bott-Samelson over $SL_3/B$ is not homeomorphic to $(\mathbb{P}^1)^3$. That is a major obstacle for the approach that Allen and VA discuss above.

There are two reduced words for the longest element of $S_3$, but the automorphism of the Dynkin diagram exchanges them. So, as an abstract variety, it makes sense to talk about the Bott-Samelson for $SL_3$. We'll call it $X$.

Let $F$ be $\mathbb{P}^2$ blown up at a point, with $\pi: F \to \mathbb{P}^2$ the blowdown. Then $X$ is a $\mathbb{P}^1$ bundle over $F$. Specifically, let $Q$ be the tautological quotient over $\mathbb{P}^2$. I believe that $X$ is $\mathbb{P}(\pi^* Q)$.

Let's compute $H^*(X)$. Everything is in even degree. $H^*(\mathbb{P}^2) = \mathbb{Z}[H]/H^3$, where $H$ is the hyperplane class. The blowup $F$ has $H^*(F) = \mathbb{Z}[H,E]/\langle H^2+E^2,\ HE \rangle$ where $E$ is the class of the exceptional fiber. The chern class of $Q$ is $1+H+H^2$. So $$H^(X) = H^*(F)[Z]/\langle Z^2 + ZH + H^2 \rangle = \mathbb{Z}[H,E,Z]/\langle H^2+E^2,\ HE,\ Z^2+ZH+H^2 \rangle.$$

So $H^2(X)$ is three dimensional. We have a cubic form on $H^2(X)$ given by $\alpha \mapsto \int_X \alpha^3$. I get that this cubic form is $$\int_X \begin{pmatrix} Z^3 & & & \\ Z^2 H & Z^2 E & & \\ Z H^2 & ZHE & Z E^2 & \\ H^3 & H^2 E & H E^2 & E^3 \end{pmatrix}= \begin{pmatrix} 0 & & & \\ 1 & 0 & & \\ -1 & 0 & 1 & \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Of course, the corresponding matrix for $(\mathbb{P}^1)^3$ is $$\begin{pmatrix} 0 & & & \\ 0 & 0 & & \\ 0 & 1 & 0 & \\ 0 & 0 & 0 & 0 \end{pmatrix}$$

Now, if there were a homeomorphism $X \cong (\mathbb{P}^1)^3$, then there would be an isomorphism $ H^2(X) \otimes \mathbb{C} \cong H^2((\mathbb{P}^1)^3) \otimes \mathbb{C}$ taking one cubic to the other. But I get that the first cubic is a line times a conic, while the second is three lines.