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Edit: The previous answer had an error, which I realized from a comment of @Will Sawin, and I've completely revised it.

This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Bass-Serre tree, hence is virtually free.

The rational quaternions is a quaternion algebra with Hilbert symbol $\binom{\underline{-1,-1}}{\mathbb{Q}}$. We may tensor with $F=\mathbb{Q}(\sqrt{5})$ to get the quaternion algebra $A=\binom{\underline{-1,-1}}{F}$. The given elements $q_i$ are unit norm elements in the quaternion algebra $A$. Since the real quaternions $\binom{\underline{-1,-1}}{\mathbb{R}}$ is ramified (i.e. a division algebra), the algebra $\binom{\underline{-1,-1}}{F}$ is ramified at both real places (tensoring $A$ with $\mathbb{R}$ over the two embeddings of $F$ into $\mathbb{R}$.

For all odd places (i.e. tensoring $A$ with $F_\mathcal{P}$, the $\mathcal{P}$-adic completion of $F$), the quaternion algebra $A_\mathcal{P}=A\otimes_F F_\mathcal{P}$ splits, i.e. is isomorphic to a matrix algebra $M_2(F_\mathcal{P})$. Since $2$ does not split over $F$, the algebra $A_{(2)}$ must also split, since $A$ must split at an even number of places by Hilbert's Reciprocity Law.

The given elements lie in an order $\mathbb{Z}[\sqrt{5}][\frac12][1,i,j,k] \subset A$. For each odd prime $\mathcal{P}$, this lies in a compact subgroup of $A_\mathcal{P}$, and lies in a compact subgroup of the real places. So it must be a lattice in $A_{(2)}$, and acts discretely cocompactly on the associated Bass-Serre tree. Thus, the group is virtually free.

The residue field of $\mathcal{O}_{F_{(2)}}$ is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Bass-Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Bass-Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this. However, it's clear that the automorphism group $A_5$ of the dodecahedron stabilizes a vertex of the Bass-Serre tree, and the twin dodecahedra should have automorphism group stabilizing adjacent vertices of the tree.

Edit: The previous answer had an error, which I realized from a comment of @Will Sawin, and I've completely revised it.

This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Serre tree associated to $SL_2$ (really, a Bruhat-Tits building associated to $SL_2(F)$, where $F$ is a local field), hence is virtually free.

The rational quaternions is a quaternion algebra with Hilbert symbol $\binom{\underline{-1,-1}}{\mathbb{Q}}$. We may tensor with $F=\mathbb{Q}(\sqrt{5})$ to get the quaternion algebra $A=\binom{\underline{-1,-1}}{F}$. The given elements $q_i$ are unit norm elements in the quaternion algebra $A$. Since the real quaternions $\binom{\underline{-1,-1}}{\mathbb{R}}$ is ramified (i.e. a division algebra), the algebra $\binom{\underline{-1,-1}}{F}$ is ramified at both real places (tensoring $A$ with $\mathbb{R}$ over the two embeddings of $F$ into $\mathbb{R}$.

For all odd places (i.e. tensoring $A$ with $F_\mathcal{P}$, the $\mathcal{P}$-adic completion of $F$), the quaternion algebra $A_\mathcal{P}=A\otimes_F F_\mathcal{P}$ splits, i.e. is isomorphic to a matrix algebra $M_2(F_\mathcal{P})$. Since $2$ does not split over $F$, the algebra $A_{(2)}$ must also split, since $A$ must split at an even number of places by Hilbert's Reciprocity Law.

The given elements lie in an order $\mathbb{Z}[\sqrt{5}][\frac12][1,i,j,k] \subset A$. For each odd prime $\mathcal{P}$, this lies in a compact subgroup of $A_\mathcal{P}$, and lies in a compact subgroup of the real places. So it must be a lattice in $A_{(2)}^1\cong SL_2(F_{(2)})$. Therefore, it acts on the tree associated to $SL_2(F_{(2)})$, described in Serre's book Trees Chapter II.1 (this is the Bruhat-Tits building associated to $SL_2(F_{(2)})$). Thus, the group is virtually free.

The residue field of $\mathcal{O}_{F_{(2)}}$ is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this. However, it's clear that the automorphism group $A_5$ of the dodecahedron stabilizes a vertex of the Serre tree, and the twin dodecahedra should have automorphism group stabilizing adjacent vertices of the tree.

This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Bass-Serre tree, hence is virtually free.

The unit quaternions $q_i$ may be interpreted as elements in $SU(2)$, treating left multiplication as unitary transformations of $\mathbb{C}^2$. The entries to the matrices lie in $\mathbb{Q}(\sqrt{5},i)$, and traces lie in $\mathbb{Q}(\sqrt{5})$. Then $tr(q_i)=-\frac12$, and one may apply Theorem 3.6.2 of MacLachlan-Reid to conclude that the quaternion algebra generated by this group over $\mathbb{Q}(\sqrt{5})$ has Hilbert symbol $(-15, -110)$. Thus, the quaternion algebra ramifies at both real places. I consulted Matt Stover at lunch, who also explained to me that it is ramified at both places sitting over the prime $11$, and no where else. So the group acts discretely on the Bass-Serre tree associated to the prime $2$ (which doesn't split in $\mathbb{Q}(\sqrt{5})$). Thus, the group is virtually free.

I'm not sure how to prove that it's freely generated by $q_i$, without explicitly computing the action on the Bass-Serre tree. One might be able to take advantage of the $S_4$ symmetry of the generators to determine this. I think the Bass-Serre tree will be defined by orders in the quaternion algebra over $\mathbb{Z}[(1+\sqrt{5})/2][\frac12]$ (or maybe more properly working over the 2-adic completion). The residue field is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Bass-Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Bass-Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this.

Edit: The previous answer had an error, which I realized from a comment of @Will Sawin, and I've completely revised it.

This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Bass-Serre tree, hence is virtually free.

The rational quaternions is a quaternion algebra with Hilbert symbol $\binom{\underline{-1,-1}}{\mathbb{Q}}$. We may tensor with $F=\mathbb{Q}(\sqrt{5})$ to get the quaternion algebra $A=\binom{\underline{-1,-1}}{F}$. The given elements $q_i$ are unit norm elements in the quaternion algebra $A$. Since the real quaternions $\binom{\underline{-1,-1}}{\mathbb{R}}$ is ramified (i.e. a division algebra), the algebra $\binom{\underline{-1,-1}}{F}$ is ramified at both real places (tensoring $A$ with $\mathbb{R}$ over the two embeddings of $F$ into $\mathbb{R}$.

For all odd places (i.e. tensoring $A$ with $F_\mathcal{P}$, the $\mathcal{P}$-adic completion of $F$), the quaternion algebra $A_\mathcal{P}=A\otimes_F F_\mathcal{P}$ splits, i.e. is isomorphic to a matrix algebra $M_2(F_\mathcal{P})$. Since $2$ does not split over $F$, the algebra $A_{(2)}$ must also split, since $A$ must split at an even number of places by Hilbert's Reciprocity Law.

The given elements lie in an order $\mathbb{Z}[\sqrt{5}][\frac12][1,i,j,k] \subset A$. For each odd prime $\mathcal{P}$, this lies in a compact subgroup of $A_\mathcal{P}$, and lies in a compact subgroup of the real places. So it must be a lattice in $A_{(2)}$, and acts discretely cocompactly on the associated Bass-Serre tree. Thus, the group is virtually free. 

The residue field of $\mathcal{O}_{F_{(2)}}$ is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Bass-Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Bass-Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this. However, it's clear that the automorphism group $A_5$ of the dodecahedron stabilizes a vertex of the Bass-Serre tree, and the twin dodecahedra should have automorphism group stabilizing adjacent vertices of the tree.

This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Bass-Serre tree, hence is virtually free.

The unit quaternions $q_i$ may be interpreted as elements in $SU(2)$, treating left multiplication as unitary transformations of $\mathbb{C}^2$. The entries to the matrices lie in $\mathbb{Q}(\sqrt{5},i)$, and traces lie in $\mathbb{Q}(\sqrt{5})$. Then $tr(q_i)=-\frac12$, and one may apply Theorem 3.6.2 of MacLachlan-Reid to conclude that the quaternion algebra generated by this group over $\mathbb{Q}(\sqrt{5})$ has Hilbert symbol $(-15, -110)$. Thus, the quaternion algebra ramifies at both real places. I consulted Matt Stover at lunch, who also explained to me that it is ramified at both places sitting over the prime $11$, and no where else. So the group acts discretely on the Bass-Serre tree associated to the prime $2$ (which doesn't split in $\mathbb{Q}(\sqrt{5})$). Thus, the group is virtually free.

I'm not sure how to prove that it's freely generated by $q_i$, without explicitly computing the action on the Bass-Serre tree. One might be able to take advantage of the $S_4$ symmetry of the generators to determine this.

This group is a subgroup of an S-arithmetic lattice, which acts discretely on finite-valence Bass-Serre tree, hence is virtually free.

The unit quaternions $q_i$ may be interpreted as elements in $SU(2)$, treating left multiplication as unitary transformations of $\mathbb{C}^2$. The entries to the matrices lie in $\mathbb{Q}(\sqrt{5},i)$, and traces lie in $\mathbb{Q}(\sqrt{5})$. Then $tr(q_i)=-\frac12$, and one may apply Theorem 3.6.2 of MacLachlan-Reid to conclude that the quaternion algebra generated by this group over $\mathbb{Q}(\sqrt{5})$ has Hilbert symbol $(-15, -110)$. Thus, the quaternion algebra ramifies at both real places. I consulted Matt Stover at lunch, who also explained to me that it is ramified at both places sitting over the prime $11$, and no where else. So the group acts discretely on the Bass-Serre tree associated to the prime $2$ (which doesn't split in $\mathbb{Q}(\sqrt{5})$). Thus, the group is virtually free.

I'm not sure how to prove that it's freely generated by $q_i$, without explicitly computing the action on the Bass-Serre tree. One might be able to take advantage of the $S_4$ symmetry of the generators to determine this. I think the Bass-Serre tree will be defined by orders in the quaternion algebra over $\mathbb{Z}[(1+\sqrt{5})/2][\frac12]$ (or maybe more properly working over the 2-adic completion). The residue field is $\mathbb{Z}[(1+\sqrt{5})/2]/(2)=\mathbb{F}_4$, the field with 4 elements, so the Bass-Serre tree has degree 5 ($=|\mathbb{P^1F}_4|$). It is tempting to guess that vertices of the Bass-Serre tree will correspond to dodecahedra, and neighbors to twins, but I haven't checked this.