If $G$ is a compact group, and we choose the normalized Haar measure $\mu$ on $G$, then $L^2(G)$ is a Banach algebra with convolution. Indeed, if $f$ and $h$ are in $L^2(G)$, then we use Hölder's inequality at the second step, together with the fact that the constant function 1 has $L^2$-norm equal to 1, to compute \begin{align*} |f\ast h(x)|\leq \int_G|f(y)h(x-y)|1\ d\mu(y)\leq \|f\|_2\|h\|_2. \end{align*} Thus $$\|f\ast h\|_2\leq \|f\ast h\|_{\infty}\leq \|f\|_2 \|h\|_2.$$ This Banach $\ast$-algebra is not only reflexive, but it is also a Hilbert space.

If $G$ is a compact group, and we choose the normalized Haar measure $\mu$ on $G$, then $L^2(G)$ is a Banach algebra with convolution. Indeed, if $f$ and $h$ are in $L^2(G)$, then we use Hölder's inequality at the second step, together with the fact that the constant function 1 has $L^2$-norm equal to 1, to compute \begin{align*} |f\ast h(x)|\leq \int_G|f(y)h(x-y)|1\ d\mu(y)\leq \|f\|_2\|h\|_2. \end{align*} Thus $$\|f\ast h\|_2\leq \|f\ast h\|_{\infty}\leq \|f\|_2 \|h\|_2.$$ This Banach $\ast$-algebra is not only reflexive, but it is also a Hilbert space.

EDIT: I was short of time yesterday when I wrote this, and in particular didn't see Ozawa's comment. I want to add that our examples are closely related. In fact, and this is based on a conversation with Chris Phillips, write $L^2(G)=\bigoplus_{\pi\in\widehat{G}}H_\pi$ according to Peter-Weyl Theorem, and give the Hilbert-Schmidt operators on $H_\pi$ the Hilbert-Schdmidt norm: $\|T\|=tr(T^*T)^{1/2}$, with composition as product, and the usual adjoint of operators on Hilbert spaces. Then I think that $L^2(G)$, with convolution as product and complex conjugation as involution, should be isomorphic to the direct sum (over $\pi$ in $\widehat{G}$) of the Hilbert-Schmidt operators on $H_\pi$.

The example I gave is, in some sense, a sum of a bunch of Ozawa's examples.

If $G$ is a compact group, and we choose the normalized Haar measure $\mu$ on $G$, then $L^2(G)$ is a Banach algebra with convolution. Indeed, if $f$ and $h$ are in $L^2(G)$, then we use Holder at the second step to compute \begin{align*} |f\ast h(x)|\leq \int_G|f(y)h(x-y)|1\ d\mu(y)\leq \|fh\|_2\|1\|_2\leq \|f\|_2\|h\|_2. \end{align*} Thus $$\|f\ast h\|_2\leq \|f\ast h\|_{\infty}\leq \|f\|_2 \|h\|_2.$$ This Banach $\ast$-algebra is not only reflexive, but it is also a Hilbert space.

If $G$ is a compact group, and we choose the normalized Haar measure $\mu$ on $G$, then $L^2(G)$ is a Banach algebra with convolution. Indeed, if $f$ and $h$ are in $L^2(G)$, then we use Hölder's inequality at the second step, together with the fact that the constant function 1 has $L^2$-norm equal to 1, to compute \begin{align*} |f\ast h(x)|\leq \int_G|f(y)h(x-y)|1\ d\mu(y)\leq \|f\|_2\|h\|_2. \end{align*} Thus $$\|f\ast h\|_2\leq \|f\ast h\|_{\infty}\leq \|f\|_2 \|h\|_2.$$ This Banach $\ast$-algebra is not only reflexive, but it is also a Hilbert space.