Something I'm a little confused on I think-- rusty on my measure theory! So I'm pretty sure I convinced myself that in general for $f \in C_c(G)$, $||f||_2 \leq ||f||_{op} \leq ||f||_1$.
However, does this imply the above nesting? I know that $L^1$ is contained in both, so it's mainly the embedding $C^*_r(G) \subset L^2(G)$ which I could do with a little explanation for!
The reason I thought this might be the case was from reading the paper "Rapidly Decreasing Functions in Reduced C^* algebras of Groups" by Jolissaint:
On the top of page 6 he gives a brief explanation, which I do not understand!
in a slightly unorthodox fashion wanted to answer my own question! Okay so basically consider the map: $$\iota:C^*_r(G) \rightarrow L^2(G) \quad a \mapsto a(\delta_1)$$ Where $a$ acts via the left regular representation. This map is clearly continuous. Considering both the left regular representation $\lambda$ and the right regular representation $\rho$ it is known that these representations commute, i.e. for all $x, y \in C^*_r(G), \; \lambda_x \rho_y=\rho_y \lambda_x$. Then, $\iota(a)=\iota(b)$ implies that $\rho_x(\iota(a))=\rho_x(\iota_b)$. In other words, $\rho_x \lambda_a \delta_1=\rho_x \lambda_b \delta_1$ using the fact the representations commute this simplifies to $\lambda_a \delta_x=\lambda_b \delta_x$. Therefore, $a=b$.
As pointed out this shows the inclusion $C^*_r(G) \subset L^2(G)$, and the other inclusion is false.