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][1]), 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  $\dim V_{{small}}^{\Gamma}>0$.

It remains to look at the classification of $3$-dimensional spherical space forms $\Gamma\backslash S^3$ and to check for each of them whether $V_{small}^\Gamma\neq0$. If $\Gamma$ is inside the standard maximal torus (i.e. $\Gamma\backslash S^3$ is a lens space), then the weight decomposition of $V_{small}$ will help, giving probably a condition on the parameters defining $\Gamma$. 


  [1]: https://publishup.uni-potsdam.de/frontdoor/index/index/docId/6723