$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$Let $k$ be the ground field. Choose a representative $n_0$ for $w_0$ in $N_G(T)(k)$.
I get an answer that is unsatisfactory in two ways. First, I have to replace $\chi_i$ by a positive multiple $\chi$, and I cannot yet identify the multiplier. (It seems likely that, if we get a non-trivial multiple as the weight, then we can just extract a root; but I do not yet see why this is so.) Second, I get that the weight is $(-w_0\chi, \chi)$, not $(\chi, -w_0\chi)$. Probably, I am just messing up something in the normalisation.
Let $\omega$ be the weight of $f$ for $B \times B$, and put $\chi = \omega(1, \cdot)$. Evaluation at $n_0$ shows that $\omega = (-w_0\chi, \chi)$, so it suffices to show that $\chi$ is a positive multiple of $\chi_i$.
For each $j$, write $U_j$ for the root group associated to $\alpha_j$. Computing in $\SL_2$ shows that there are a representative $n_j$ in $N_G(T)(k)$ of $s_j$, and functions $u_\ell, u_r : \GL_1 \to U_j$, such that $\lim_{t \to 0} u_\ell(t)n_j\alpha_j^\vee(t)u_r(t) = 1$. Then $g_j(t) \mathrel{:=} n_0 n_j^{-1} u_\ell(t)\alpha_j^\vee(t)^{-1}n_j u_r(t)$ lies in $B w_0 B$, so that $f(g_j(t))$ is non-$0$, for all $t$, and $\lim_{t \to 0} g_j(t) = n_0 n_j^{-1}$ belongs to $B w_0 s_j B$, so that $\lim_{t \to 0} f(g_j(t))$ is $0$ exactly if $j = i$.
We have that $$ g_j(t) = U_\ell(t)^{-1}g_j(1)u_r(1)^{-1}\alpha_j^\vee(t)u_r(t), $$ where $U_\ell(t) = n_0 n_j^{-1}u_\ell(1)u_\ell(t)^{-1}n_j n_0^{-1}$, so $$ f(g_j(t)) = \omega(U_\ell(t), u_r(1)^{-1}\alpha_j^\vee(t)u_r(t))f(g_j(1)) = \chi(\alpha_j^\vee(t))f(g_j(1)) = t^{\langle\chi, \alpha_j^\vee\rangle}f(g_j(1)), $$ for all $t$. Since $\lim_{t \to 0} f(g_j(t))$ exists, we have that $\langle\chi, \alpha_j^\vee\rangle$ is non-negative; and $\langle\chi, \alpha_j^\vee\rangle$ is positive if and only if $\lim_{t \to 0} f(g_j(t)) = 0$, which happens if and only if $j = i$.