An algebra with a strictly positive state + condition ? to become a C$^*$-algebra.
No Hilbert space is a $C^*$-algebra in any natural way since the structure of Hilbert spaces (a vector space with an inner product) is quite different from that of a C*-algebra (a Banach space equipped with an involutive algebra structure).
Nevertheless from the data provided by the OP one may give $A$ the structure of a $C^*$-algebra in a very natural way. The first step is to consider the hermitian form $g$ as an inner product on $A$, so that $$ H:= (A,g) $$ becomes a Hilbert space (completeness follows from finite dimensionality). One may then define a map $$ \pi :A\to B(H), $$ by $$ \pi (a)\xi = a\xi , \quad\forall a\in A,\quad \forall \xi \in H, $$ (here $a\xi $ is nothing other than the product of $a$ and $\xi $).
One may easily show that $\pi $ is a *-homomorphism. It is moreover injective since, for $a\neq 0$, one has that $\pi (a)a^*=aa^*\neq 0$, as a consequence of $\varphi (aa^*)>0$.
Finally, identifying $A$ with its image within $B(H)$ via $\pi $, we have that $A$ becomes a $C^*$-algebra.
Later edit: The late edit indeed makes the question trivial and, as noted my Martin, the only example is the complex numbers.
A perhaps more elementary proof of this fact is that every finite dimensional $C^*$-algebra is the direct sum of matrix algebras, so unless $\text{dim}(A)=1$, there exists a pair of mutually orthogonal projections $p$ and $q$. It follows that $\|p\pm q\|=1$, hence the parallelogram law fails.
The following statements are equivalent:
$A$ is a C$^*$-algebra with the norm induced by $\varphi$
$\dim A=1$, i.e., $A=\mathbb C$
Proof. if $A=\mathbb C$, then the only state is the identity, which induces the norm.
Conversely, if $A$ is a C$^*$-algebra, its norm satisfies the C$^*$-identity $$\tag1\|x\|^2=\|x^*x\|.$$ In terms of $\varphi$, this is $$\tag2\varphi(x^*x)=\varphi((x^*x)^2)^{1/2},\qquad x\in A.$$ So, for each positive $a\in A$, $$\tag3\varphi(a)^2=\varphi(a^2).$$ As $\varphi$ is a ucp map, the equality $(3)$ says that $a$ is in the multiplicative domain of $\varphi$. Thus the multiplicative domain of $\varphi$ is all of $A$, since a C$^*$-algebra is spanned by its positive elements. So $\varphi$ is a faithful representation of $A$ into (thus onto) $\mathbb C$.