EDIT II: Sorry to bump this again but the answer to this question can be phrased entirely in terms of finite Coxeter groups and doesn’t depend at all on the fact that we’re dealing with $\textsf{E}_6$, so it seemed best to write the answer that way.

Let $V$ be a finite dimensional real euclidean vector space and let $\Phi \subseteq V$ be a root system in the sense of §1.2 in Humphrey’s “Reflection groups and Coxeter groups”. For convenience we will assume that $\Phi$ spans $V$. We will denote by $W = \langle s_{\alpha} \mid \alpha \in \Phi \rangle$ the corresponding reflection group and we choose a set of simple roots $\Delta \subseteq \Phi$.

If $w_0 \in W$ is the longest element, with respect to $\Delta$, then $-w_0(\Delta) = \Delta$ and hence $-w_0$ defines an automorphism $\sigma : V \to V$ such that $\sigma(\Delta) = \Delta$.

We will denote by $\mathcal{S}$ the set of all 1-dimensional subspaces $A \subseteq V$ such that $w(v) = -v$ for some $w \in W$ and any $v \in A$. Clearly $W$ acts on the set $\mathcal{S}$ because we have

\begin{equation*}
w(v) = -v \Leftrightarrow xwx^{-1}(x(v)) = -x(v)
\end{equation*}

for any $x \in W$. Let $D = \{\lambda \in V \mid (\lambda,\alpha) \geqslant 0$ for all $\alpha \in \Delta\}$ be the closure of the fundamental Weyl chamber of $W$, where $(-,-) : V \times V \to \mathbb{R}$ is the euclidean form on $V$. As any element in $V$ is conjugate under $W$ to a unique element in $D$, i.e., $D$ is a fundamental domain for the action of $W$, it suffices to determine for which $v \in D$ we have $\mathbb{R}v \in \mathcal{S}$.

Assume $v \in D$ is such that $w(v) = -v$ for some $w \in W$ then we have

\begin{equation*}
w_0w(v) = -w_0(v) = \sigma(v).
\end{equation*}

Now assume $\lambda \in D$ then $(\sigma(\lambda),\sigma(\alpha)) = (\lambda,\alpha) \geqslant 0$ for all $\alpha \in \Delta$. As $\sigma(\Delta) = \Delta$ this shows that $\sigma(D) = D$. Hence $w_0w(v) \in D$ but as $v \in D$ this implies $w_0w(v) = v$ and $w_0w = 1$. In particular, we have $\sigma(v) = v$ and $w = w_0$.

Consider the subgroup $H = \{w \in W \mid w_0ww_0 = w\}$ fixed under conjugation by $w_0$. If $W/H$ are the cosets of $H$ in $W$ then it is clear from the above discussion that we have

\begin{equation*}
\mathcal{S} = \bigsqcup_{x \in W/H} x(\mathcal{S}^{\sigma})
\end{equation*}

where $\mathcal{S}^{\sigma} = \{A \in \mathcal{S} \mid \sigma(A) = A\}$.

Firstly, if $w_0 = -1$ then it is clear that $\mathcal{S}$ simply contains all 1-dimensional subspaces of $V$. Now, let us consider your special situation of $\textsf{E}_6$. We number the simple roots $\Delta = \{\alpha_1,\dots,\alpha_6\}$ as in Bourbaki. In this case $\sigma(\alpha_i) = \alpha_{\rho(i)}$ where $\rho \in \mathfrak{S}_6$ is the permutation $(1,6)(3,5)$. It is well known that the subgroup $H$ is isomorphic to the Weyl group $W_{\textsf{F}_4}$, see for instance 13.3.3 of Carter's "Simple groups of Lie type". Comparing the orders of the Weyl groups of type $\textsf{E}_6$ and $\textsf{F}_4$ we see that $|W/H| = 45$, which explains your 45 tori.

I am not aware of a reference for this fact but in the end it's reasonably straight forward.