This is an answer to Q2 that has nothing to do with interpolation. Here I am not trying to give general criteria for $\|T\|<1$, as this is a hard question, only to relate this "spectral gap" property for different $p$'s. Following the (ongoing) conversation in the comments thread, I will assume for start that the measure is finitely supported (which is indeed easier, as Mikael indicates, and already interesting).
Let $G$ be a compact group, $\mu$ a finitely supported probability measure on $G$. Let $T_p:L^0_p(G)\to L^0_p(G)$ be the operator given by left convolution with $\mu$.
Claim: Either for every $1<p<\infty$, $\|T_p\|=1$ or for every $1<p<\infty$, $\|T_p\|<1$.
Let $\Gamma<G$ be the (countable) group generated by the support of $\mu$. We assume as we may that $\Gamma$ is dense in $G$, otherwise $\|T_p\|=1$ for all $p$. It follows that there are no $\Gamma$-invariant vectors. By strict convexity of the norm we conclude that there are no $T_p$-invariant vectors. Moreover, by the uniform convexity of the norm it is not hard to see the equivalence (for a given $p$): $T_p$ has almost invariant vectors iff $\Gamma$ has almost invariant vectors. The former is equivalent to $\|T_p\|=1$. However the latter is independent of $p\in (1,\infty)$. Indeed, for every $p,q$ we can define the (non-linear) Mazur map $L_p\to L_q$, $f\mapsto \text{sgn}(f)|f|^{p/q}$ which has two nice properties: it is uniformly continuous on the sphere and it commutes with isometries, ie with $\Gamma$ (this is for $p\neq 2$, by Banach-Lamperti). For more details see section 4 in http://arxiv.org/pdf/math/0506361v2.pdf.
It follows that the statement $\|T_p\|=1$ is independent of $p$ which proves the claim.