Here is more direct and elementary argument.

**Lemma:** $\mathrm{Aut}(W_3)$ and $\mathrm{Aut}(\mathbb{F}_2)$ are isomorphic, where $W_3$ denotes the free product $\mathbb{Z}_2 \ast \mathbb{Z}_2 \ast \mathbb{Z}_2$.

*Sketch of proof.* Let $F \leq W_3$ denote the kernel of the morphism $W_3 \twoheadrightarrow \mathbb{Z}_2$ sending all the generators $a,b,c$ to $1$. In other words, $F$ coincides with the elements that can be written as reduced words of even lengths. Because $F$ is a characteristic subgroup freely generated by $\{ab,bc\}$, we find a morphism $\mathrm{Aut}(W_3) \to \mathrm{Aut}(\mathbb{F}_2)$. It can be checked that it is an isomorphism. $\square$

I found this statement in Varghese's article *The automorphism group of the universal Coxeter group*. In the sequel, I work with $\mathrm{Aut}(W_3)$ instead of $\mathrm{Aut}(\mathbb{F}_2)$, but it is just because I am more confortable with it. As the isomorphism above is completely explicit, you can translate everything in $\mathrm{Aut}(\mathbb{F}_2)$ if you want.

**Proposition:** $\mathrm{Aut}(W_3)$ is not hyperbolic relative to proper subgroups.

*Proof.* Fix a basis $\{a,b,c\}$ of $W_3$. For every $g \in W_3$, let $\iota_g$ denote the inner automorphism given by the conjugation by $g$. Also, set
$$\kappa_1 : \left\{ \begin{array}{ccc} a & \mapsto & a \\ b & \mapsto & b \\ c & \mapsto & c^{ab} \end{array} \right. \text{ and } \kappa_2 : \left\{ \begin{array}{ccc} a & \mapsto & a^{cb} \\ b & \mapsto & b \\ c & \mapsto & c \end{array} \right..$$
They are partial conjugations with infinite-order images in $\mathrm{Out}(W_3)$. (I use the notation $g^h=hgh^{-1}$.) Finally, set
$$\varphi := \iota_{ab} \circ \kappa_1, \ \psi := \iota_b \circ \varphi \circ \iota_b, \text{ and } \xi := \iota_{cb} \circ \kappa_2.$$
The key observation is that $\varphi$ (resp. $\psi$, $\xi$) fixes $ab$ and $bcb$ (resp. $ab$ and $c$, $bc$ and $bab$). As a consequence, the subgroups
$$\begin{array}{l} A:= \langle \iota_{ab}, \iota_{bcb}, \varphi \rangle \\ B:= \langle \iota_{ab}, \iota_c, \psi \rangle \\ C:=\langle \iota_{bc}, \iota_{bab}, \xi \rangle \end{array}$$
split as direct products between a(n infinite) virtually free group (namely, the part in the inner subgroup) and an infinite cyclic group. This implies that, if we assume that $\mathrm{Aut}(W_3)$ is hyperbolic relative to some collection of subgroups $\mathcal{P}$, then there exist $I,J,K \in \mathcal{P}$ such that $A \subset I$, $B \subset J$, and $C \subset K$. Now, we can conclude because $\mathcal{P}$ must be an almost malnormal collection of subgroups. Indeed:

- Because $I \cap J \supset \langle \iota_{ab} \rangle$, necessarily $I=J$.
- This implies that $J \supset \langle \iota_{ab}, \iota_{bcb}, \iota_c \rangle \supset \langle \iota_{(bc)^2} \rangle$, hence $J \cap K \supset \langle \iota_{(bc)^2} \rangle$, and finally $J=K$.
- Thus, $I=J=K$ contains $\langle \iota_{ab} , \iota_{bcb}, \iota_{c}, \iota_{bc}, \iota_{bab} \rangle= \mathrm{Inn}(W_3)$.
- Finally, from the fact that $\mathrm{Inn}(W_3)$ is a normal subgroup in $\mathrm{Aut}(W_3)$, we conclude that $I= \mathrm{Aut}(W_3)$.

Thus, we have proved that $\mathcal{P}$ must contain a subgroup that is not proper. $\square$