Does there exist a nontrivial (i.e. other than $\mathbb{C}$) example of a $C^*$-algebra which is also a Hilbert space (in the same norm, of course)?
For $\mathbb{C}^n$ with $n > 1$ the answer is no by uniqueness of norm in $C^*$-algebras, since $\mathbb{C}^n$ is a $C^*$-algebra in the $\ell^\infty$ norm, which is not given by an inner product. What about more generally?
No. Every infinite dimensional C*-algebra is a strict subset of its second dual.
Edited to add: Perhaps an easier way to see it is to use the fact that the extreme points of the unit ball of a C*-algebra are precisely the partial isometries. It is not too hard to see that any C*-algebra of dimension greater than 1 contains elements of norm one that are not partial isometries.
Edit the second: Just can't resist giving one more reason. First, the unit ball of a non-unital C*-algebra has no extreme points at all. Otherwise, call the unit $e$ and the cone of positive elements $P$, then the unit ball in the self-adjoint part of the C*-algebra is $B_{\text{sa}}=(e-P)\cap(-e+P)$. In particular, $B_{\text{sa}}\subset(e-P)$. It is easy to that if this holds for a unit vector $e$ in a Hilbert space, the cone $P$ must be the entire half space $\{v\colon \langle v,e\rangle\ge0\}$. In geometric terms, the unit ball of the self-adjoint part of a C*-algebra has a sharp point at the unit, while the unit ball of a Hilbert space is smooth.
I will first assume that the $C^*$-algebra is unital. If it is not one-dimensional, then because every element of the algebra has the form $a+ib$ with $a$ and $b$ self-adjoint, there is a self-adjoint element $a$ that is not a scalar multiple of the identity. Then the spectrum of $a$ contains at least $2$ distinct elements. Otherwise, if say $\lambda$ were the only element of the spectrum of $a$, then $a-\lambda 1$ would be a self-adjoint element with spectral radius $0$, hence $a=\lambda 1$.
Since the spectrum of $a$ has at least $2$ elements, there exist continuous ($\mathbb C$-valued or even positive real-valued) functions $f$ and $g$ on the spectrum of $a$ such that $\|f(a)\|=\|g(a)\|=1$ and $f(a)g(a)=0$. It follows that $\|f(a)+g(a)\|=\|f(a)-g(a)\|=1$. Hence
$$2\|f(a)\|^2+2\|g(a)\|^2=4>2=\|f(a)+g(a)\|^2+\|f(a)-g(a)\|^2,$$ in violation of the parallelogram law. Alternatively, $\{f(a)+tg(a):t\in[-1,1]\}$ is a closed convex set with infinitely many elements of minimal norm.
If the algebra is nonunital, you can still find $a$, $f$, and $g$ as above, but you need $f(0)=g(0)=0$, so the spectrum of $a$ should have $2$ distinct nonzero elements, which means slightly more is needed to see that such $a$ exists. This feels like overkill, but in this case the algebra is infinite dimensional, which implies that it has a self-adjoint element with infinite spectrum. Note that a self-adjoint element whose spectrum has only one nonzero element is a scalar multiple of a projection, and there should be an easier way to see that there are self-adjoint elements that are not multiples of projections, but none comes to mind.
