Remember that $S^3=SO(4)/SO(3)=:G/K$ as a symmetric space.
The following actually holds for every compact symmetric space $M=G/K$.
Take the Cartan decomposition $\mathfrak g=\mathfrak k\oplus \mathfrak p$.
By using that the vector bundle $\mathcal S_0^2(M)$ of traceless symmetric 2-forms is homogeneous (i.e. $\mathcal S_0^2(M)\simeq G\times_{\tau} Sym^2(\mathcal m^{\mathbb C})$, where $\tau$ denotes the $K$-action on $Sym^2(\mathcal m^{\mathbb C})$), it decomposes as
$$
\mathcal S_0^2(M)^{\mathbb C} = \widehat{\bigoplus}_{\pi\in\widehat {G}} V_\pi\otimes Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C}).
$$
Of course, this sum is reduced to those $\pi\in\widehat G$ such that $Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C})\neq0$.
Now, since you are interested on $\Gamma\backslash S^3 = \Gamma\backslash G/K$ with $\Gamma$ a discrete subgroup of $G$, one has that
$$
\mathcal S_0^2(\Gamma\backslash M)^{\mathbb C} = \widehat{\bigoplus}_{\pi\in\widehat {G}} V_\pi^{\Gamma}\otimes Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C}).
$$
In conclusion, the sum is now restricted to those $\pi\in\widehat G$ such that $Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C})\neq0$ and $V_\pi^{\Gamma}\neq 0$.
The Lichnerowicz Laplacian on $\mathcal S_0^2(M)$ (or $\mathcal S_0^2(\Gamma\backslash M)$) acts by the casimir element (here, it is crucial that $G/K$ is symmetric), so each term $V_\pi^{\Gamma}\otimes Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C})$ contributes to the spectrum of the Lichnerowicz Laplacian with the eigenvalue $\lambda_\pi$ such that $Cas\cdot V_\pi = \lambda_\pi Id_{V_\pi}$ (which can be computed with Freudenthal's formula) with multiplicity $\dim V_\pi^{\Gamma}\dim Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C})$.
Let $\lambda_1(M)$ denote the smallest eigenvalue of the Lichnerowicz Laplacian $\Delta_L$ on $M$.
(Since $\Delta_L h=\Delta h-6h$ for spherical space forms, where $\Delta$ is the Rough Laplacian, then it is equivalent to work with any of them).
One has
$$
\lambda_1(G/K) =\min_\pi \lambda_\pi,
$$
where $\pi$ runs over $\widehat G$ such that $\dim Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C})>0$, and at least one element in $\mathcal S_0^2(M)$ induced by $V_\pi\otimes Hom_K(V_\pi,Sym_0^2(\mathfrak p)^{\mathbb C})$ is divergence-free (note that I haven't mentioned this condition before).
Concerning your question about whether $\lambda_1(\Gamma\backslash G/K)=\lambda_1(G/K)$ for a given $\Gamma\subset G$, it reduces to check whether $\dim V_\pi^\Gamma>0$ for any $\pi$ attaining the displayed identity above, that is, $\lambda_1(G/K)=\lambda_\pi$.
For the sphere $G/K=S^3$, one has that $\lambda_1(S^3)=12$ (see for instance Example 3.1.2 in Klaus Kröncke's PhD thesis), and it is attainded only by $\pi_{small}:=\pi_{2\varepsilon_1+2\varepsilon_2}$ (the irrep of $SO(4)$ having highest weight $2\varepsilon_1+2\varepsilon_2$).
In conclusion, $\lambda_1(\Gamma\backslash S^3)=12$ if and only if $\dim V_{{small}}^{\Gamma}>0$.
Now, the original problem of classifying 3-spherical space forms $\Gamma\backslash S^3$ satisfying $\lambda_1(\Gamma\backslash S^3)=12$ reduces to the following:
Classify the finite subgroups $\Gamma\subset SO(4)$ acting freely on
$S^3$ such that $V_{small}^\Gamma\neq 0$.
For instance, if $\Gamma$ is cyclic (hence $\Gamma\backslash S^3$ is a lens space), it is conjugate in $O(4)$ to the group generated by
$$
\begin{pmatrix}
\cos(2\pi /q) & \sin(2\pi /q)\\
-\sin(2\pi /q) & \cos(2\pi /q) \\
&& \cos(2\pi p/q) & \sin(2\pi p/q)\\
&& -\sin(2\pi p/q)& \cos(2\pi p/q)\\
\end{pmatrix}
$$
for some $q\in\mathbb N$ and $p\in\mathbb Z$ coprime to $q$.
The lens space $\Gamma\backslash S^3$ is usually denoted by $L(q;1,p)$ or just $L(q,p)$.
One can check that $V_{small}^\Gamma\neq0$ if and only if $q$ divides $2(1+p)$ or $2(1-p)$. In particular, $\lambda_1(L(q;1,1))=12$ for all $q\in\mathbb N$.
The calculation involved for the non-cyclic subgroups of $SO(4)$ are much more complicated.
