What is going on in this degree 8 number field that fails to be a quaternion extension of $\mathbb{Q}$?

I figured it out.

First of all, we can think of $\alpha$ as $\sqrt{\sqrt{p}+\sqrt{q}}$, with $p,q$ rationals such that $-4pq=1$. Then $r=p+q$, $\psi = \sqrt{p-q}$. Also, $i=2\sqrt{pq}$, and $\sqrt{2(r-\psi^2)}=2\sqrt{q}$, explaining why $i,\sqrt{2(r-\psi^2)}$ are in $\mathbb{Q}(\alpha^2)$.

This makes it clear that, in general, $\psi\notin L$.

Then, $f$'s splitting field is $L(\psi)$; an automorphism of this field is determined by its action on $\alpha,\psi$. Defining $\mathbf{i},\mathbf{j}$ as in the question, and setting $\mathbf{k}=\mathbf{ij}$, we find that $\mathbf{k}$ sends $\alpha\mapsto i\alpha$ and $\psi\mapsto -\psi$. Let $\mathbf{-1}$ denote the automorphism that fixes $\psi$ and sends $\alpha\mapsto -\alpha$; it commutes with the others. Then $\mathbf{i},\mathbf{j},\mathbf{k}$ satisfy the famous $\mathbf{i}^2=\mathbf{j}^2=\mathbf{k}^2=\mathbf{ijk}=\mathbf{-1}$. Thus $G=\operatorname{Gal}(L(\psi)/\mathbb{Q})$ does contain $Q_8$ as a subgroup.

I was secretly hoping that $G$ would also have $Q_8$ as a quotient, so I could still pull a quaternion extension of $\mathbb{Q}$ out of all this, but alas, this is not to be. $G$ is generated over $Q_8$ by a central element $\mathbf{l}$, that, tragically, has order 4. (For a moment, I though it might have order $2$ and thus imply that $G=Q_8\times C_2$. Alas...) We can find $\mathbf{l}$ as follows:

Let $\sigma$ be the automorphism of $L(\psi)$ that sends $\psi\mapsto -\psi$ and fixes $\alpha$, i.e. the nontrivial automorphism of $L(\psi)$ over $L$. It is clear that $\sigma$ generates $G$ over $Q_8$ because $Q_8$ is index $2$ and $\sigma\notin Q_8$.

Now $L(\psi)$ has $\mathbb{Q}(\sqrt{p},\sqrt{q},\psi)$ as a subfield. Restriction to this subfield gives us a surjective homomorphism $G\rightarrow \operatorname{Gal}(\mathbb{Q}(\sqrt{p},\sqrt{q},\psi))$; thus we have an exact sequence

$$1\rightarrow C_2\rightarrow G\rightarrow C_2\times C_2\times C_2\rightarrow 1$$

A normal subgroup of order $2$ is automatically central, thus $G$ is a central extension of $C_2\times C_2\times C_2$ by $C_2$. It was pointed out to me by Fedor Bogomolov that whenever $A,B$ are abelian groups and $G$ is a central extension of $A$ by $B$, the commutator $[\cdot,\cdot]$ induces a $\mathbb{Z}$-bilinear form on $A$ with values in $B$. (This is awesome! This must be a standard fact, right? Where would one normally learn this?) This is because commutators $[x,y]$ in $G$ are necessarily in $B$ because they are knocked out by homomorphing to $A$ since it is abelian, and furthermore they are not affected by changing $x,y$ by central elements (in particular, by elements of $B\subset Z(G)$), thus $[\cdot,\cdot]$ is still unambiguous when it is seen as accepting values from $A$ rather than $G$.

In the present case, $A=C_2\times C_2\times C_2$ is an $\mathbb{F}_2$ vector space and $B=C_2=\mathbb{F}_2$, so we can see the commutator as a skew form on $\mathbb{F}_2^3$!

Any skew form on an odd-dimensional space has a nontrivial null space. The preimage in $G$ of this nullspace is the center of $G$. This gives us a way of finding a central element of $G$ that generates it over $Q_8$.

Let $\bar{\mathbf{i}},\bar{\mathbf{k}},\bar{\sigma}$ be the images of $\mathbf{i},\mathbf{k},\sigma$ in $C_2\times C_2\times C_2$. $\mathbf{i}$ reverses $\psi$ and $\sqrt{q}$ while fixing $\sqrt{p}$; $\mathbf{k}$ reverses all three; while $\sigma$ just reverses $\psi$. Thus $\bar{\mathbf{i}},\bar{\mathbf{k}},\bar{\sigma}$ are a basis of $C_2\times C_2 \times C_2$. Since $\mathbf{k},\sigma$ commute while $\mathbf{i}$ doesn't commute with either, the matrix of the commutator form with respect to this basis is

$$\begin{pmatrix}0&1&1\\1&0&0\\1&0&0\end{pmatrix}$$

and, by inspection, the nullspace is spanned by $\begin{pmatrix}0&1&1\end{pmatrix}^T$. It follows that $\mathbf{k}\sigma$, the automorphism of $L(\psi)$ that fixes $\psi$ and sends $\alpha\mapsto i\alpha$, is in the center of $G$! Let $\mathbf{l}=\mathbf{k}\sigma$.

Then $G=Q_8\cup\mathbf{l}Q_8$, $\mathbf{l}^2=\mathbf{-1}$, and $\mathbf{l}$ commutes with everything. This is enough information to calculate in $G$.