Having worked a bit on the question, I am now able to provide an answer. It is technical and not so elegant, so I would sill be glad to see better answers, but I think it is rather elementary. The idea is to compute the equivariant Thom class of the bundle, and to restrict it to the zero-section. Since the induced bundle over $BT$ admits a $\mathbb{C}^\ast$ action, it is orientable, so it admist a Thom class, and we will see that there is only one class in $H^2_T(\mathbb{C},\mathbb{C}^\ast)$, so it will have to be the Thom class.

So let us compute $H^2_T(\mathbb{C},\mathbb{C}^\ast)$. We will proced through the equivariant long exact sequence of the pair $(\mathbb{C},\mathbb{C}^\ast)$ :
$$
\dots \to H^1_T(\mathbb{C}^\ast) \to H^2_T(\mathbb{C},\mathbb{C}^\ast) \stackrel{j^\ast}{\to} H^2_T(\mathbb{C}) \stackrel{i^\ast}{\to} H^2_T(\mathbb{C}^\ast) \to \cdots
$$
Let us write $(\mu^i) $ for a basis of $T^\ast$, the dual of the Lie algebra of $T$.
Now use any way you like to see that $H^1_T(\mathbb{C}^\ast)=0$ and $H^2_T(\mathbb{C}^\ast)=\bigoplus_i \mathbb{R}\mu^i / \alpha^*$.
So $j^\ast$ is injective, so $H^2_T(\mathbb{C},\mathbb{C}^\ast) = \ker i^\ast$. Now $\mathbb{C}$ is ($T$-equivariantly) homotopic to a point, so $H^2_T(\mathbb{C}) = \bigoplus_i \mathbb{R}\mu^i$, and $i^\ast$ simply maps $\mu^i$ to its class modulo $\alpha^\ast$, so $H^2_T(\mathbb{C},\mathbb{C}^\ast) = \ker i^\ast = \mathbb{R}\alpha^\ast$ and we have what we wanted.