From a physical viewpoint, I would dare to say that a metric must be a (symm. cov.) tensor field of rank 2 because of the principle of equivalence! But to understand why I said that, lets see the mathematics first. *By definition,* a metric on a (pseudo)-Riemannian manifold should be a symmetric covariant tensor field, and the reason is purely mathematical. Indeed, in its most elementary form, the notion of a metric arises in mathematics when we have an abstract set \(X\) together with a mapping \((p,q)\in X \times X\mapsto d(p,q)\in \mathbb R\) that satisfy the obvious conditions of what we expect to be a "distance" between the points \(p,q \in X\) (namely, \(d(p,p)=0\), symmetry \(d(x,y)=d(y,x)\) and the triangle inequality; I have not required it to be positive, so mathematicians call it a pseudo-metric). In physics, one is usually interested in spaces \(X\) that carry more structure, for example, that it be a vector space. The metric structure on such spaces should somehow be related to its linear structure. The obvious way is to introduce an inner product \((p,q)\in X \mapsto g(p,q)\), a symmetric nondegenerated bilinear form on \(X\), define a (squared) norm \(||p||^2=g(p,p)\) and declare \(d(p,q)^2=||p-q||^2\). So the metric structure derives from the linear structure.

In Euclidean spaces, of course, there is no problem in taking the square root since the inner product \(g\) is positive definite. But in relativity, we are interested in the case that the inner product is of Lorentz signature, so it is better to work with the squared norm, which is actually called the *quadratic form*. So the famous *\(ds^2\) *that appears in physics textbooks is in reality a quadratic form sending a vector \(p \in X \mapsto ds^2(p)=g(p,p)\). See for example O'Neill, *Semi-Riemannian Geometry*.

In special relativity, we can identify the spacetime manifold with its own tangent space, so Minkowski space is just the vector space \(M=\mathbb R ^4\) with the Lorentz inner product \(g(x,y)=\eta_{ab}x^ay^b\forall x,y\in M\). Now, in a general spacetime manifold, the principle of equivalence, which says that locally the spacetime should look like Minkowski, translates into the framework of differential geometry as saying that the tangent space at each spacetime point (or event) is isomorphic to Minkowski space. (Recall that in DG, infinitesimal objects goes to tangent space!) So, a metric on a general spacetime \(\mathcal M\) should be an assignment to each event \(p \in \mathcal M\) of an inner product on the tangent vector space \(T_p \mathcal M\), or: \(p\in \mathcal M \mapsto^g g_p:T_p(\mathcal M)\times T_p(\mathcal M) \longrightarrow \mathbb R\), and this precisely symm. cov. tensor field \(g \in \mathrm{sec} T^{(0,2)} \mathcal M\).

All this having being said, I think that there is simply no sense in saying that a higher order tensor (namely, higher than 2) could represent a metric. However, if what one is interested in is in constructing the metric tensor from a higher order one, than the question and therefore the answer is completely different. For example, you naturally construct the metric from the curvature tensor if you assume the curvature to be that derived from the Levi-Civita connection of the metric. But observe that I can give you a curvature without giving you a metric at all. It could be the curvature derived from a teleparallel connection, for example.