I have two, and perhaps infinitely many, examples in finite dimension $n$.
n=2. Take $X={\mathbb R}^2$ with $\ell^1$-norm $$\|x\|_1=|x_1|+|x_2|.$$ Then $X^*={\mathbb R}^2$ has the $\ell^\infty$-norm $$\|y\|_\infty=\max(|y_1|,|y_2|).$$ I turns out that $$\|x\|_1=\max(|x_1+x_2|,|x_1-x_2|)$$ and thus $X'$ is isometric to $X$, via $x\mapsto(x_1+x_2,x_1-x_2)$.
More generally, suppose that in $\mathbb R^n$, we have a convex polytope $T$ that is self-dual and is symmetric under $x\leftrightarrow-x$. Let $\|\cdot\|_T$ be the gauge associated with $T$. Then $X=(\mathbb R^n, \|\cdot\|_T)$ is isometric to $X'$ because $T$ is the unit ball of $X$ and $T'=T$ is that of $X'$.
For instance, if n=4, the polyoctahedron (= octaplex) has these properties, thus there is an $\mathbb R^4$ that is isometric to its dual, yet is not Hilbert. If $n\ge3$, the simplex is self-dual but not centro-symmetric.
This raises two questions:
Does there exist other centro-symmetric self dual convex polytopes? Maybe there exist one in any even dimension ...
Is it possible to deform the examples above so as to replace the polygone/-tope by a ball with a smooth boundary?