The $\nabla$-compatible metrics on $E$ are the positive-definite $\nabla'$-parallel sections of $S^2(E^*)$, where $\nabla'$ is the connection on $S^2(E^*)$ induced by $\nabla$. When $M$ is connected and $x\in M$ is fixed, the space of $\nabla'$-parallel sections of $S^2(E^*)$ is isomorphic to the set $Q_x(\nabla)\subset S^2(E^*_x)$ of quadratic forms on $E_x$ that are invariant under the holonomy group $\mathrm{Hol}_x(\nabla)\subset \mathrm{GL}(E_x)$.
A priori, the holonomy group $\mathrm{Hol}_x(\nabla)$ could be just about any subgroup of $\mathrm{GL}(E_x)$, so your question is really a linear algebra question of how many quadratic forms on a vector space $V$ are invariant under a given subgroup $H\subset\mathrm{GL}(V)$. The subset $Q^+_x(\nabla)\subset Q_x(\nabla)$ consisting of positive definite quadratic forms is a (possibly empty) open subset of $Q_x(\nabla)$, so if it's not empty, then it has the same dimension as $Q_x(\nabla)$. Note that $Q^+_x(\nabla)$ is non-empty if and only if the closure of $\mathrm{Hol}_x(\nabla)$ in $\mathrm{GL}(E_x)$ is compact.
The main restriction on $\mathrm{Hol}_x(\nabla)$ (when $\dim M \ge 2$) is that there has to be a surjective homomorphism
$$
\pi_1(M,x)\to \mathrm{Hol}_x(\nabla)/\mathrm{Hol}^0_x(\nabla),
$$
where $\mathrm{Hol}^0_x(\nabla)\subset \mathrm{Hol}_x(\nabla)$ is the identity component of the group $\mathrm{Hol}_x(\nabla)$. [Of course, when $M\simeq \mathbb{R}$, the group $\mathrm{Hol}_x(\nabla)$ is trivial, so $Q_x=S^2(E^*_x)$,
and when $M\simeq S^1$, the group $\mathrm{Hol}_x(\nabla)$ is the set of powers of a single element $h\in \mathrm{GL}(E_x)$.]
In any case, the dimension of $Q_x(\nabla)$ can be as low as $0$ and as high as $\tfrac12r(r{+}1)$, where $r$ is the rank of $E$ (i.e., the dimension of $E_x$).
