Can I use a higher rank tensor field as the metric field?
I) OP is interested in totally symmetric covariant $(0,r)$ tensor fields $$g\in\Gamma\left({\rm Sym}^r(T^{\ast}M)\right) \tag{A}$$ on an $n$-dimensional manifold $M$. The number of totally symmetric tensor components are $$\begin{pmatrix} n+r-1 \cr r\end{pmatrix} .\tag{B}$$
II) If the manifold $M$ is paracompact, we can use the partition of unity to prove
that there exist globally defined positive definite$^{\dagger}$ tensor fields (A).
that there exist globally defined torsion-free tangent-bundle connections $\nabla$.
(To see point 2, use point 1 for the case $r=2$ to deduce the existence of a globally defined positive definite metric tensor field, and hence a globally defined Levi-Civita connection.)
III) Next we extract the interesting part of OP's question as follows:
Can we choose a torsionfree tangent bundle connection $\nabla$ that is compatible $$\nabla g~=~0 \tag{C}$$ with a given tensor field (A)?
Generically the answer is No, not even locally, if the rank $r\geq 3$. This is because the number $$n\begin{pmatrix} n+r-1 \cr r\end{pmatrix} \tag{D}$$ of compatibility conditions (C) is greater than the number $$n \begin{pmatrix} n+1 \cr 2\end{pmatrix} \tag{E}$$ of Christoffel symbols, if $r\geq 3$. So the equations are overconstrained.
IV) We leave it to the reader to generalize the above to (not necessarily totally symmetric) higher-rank $(s,r)$ tensor fields. Higher-rank tensor fields appear e.g. in string theory, AKSZ sigma models & higher spin theories. For generalizations of Riemannian geometry, see also Finsler geometry.
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$^{\dagger}$ We define that the tensor field $g$ is positive definite if $$\forall p\in M ~\forall X_p\in T_pM\backslash\{0\}: \quad g_p(\underbrace{X_p,\ldots, X_p}_{r\text{ entries}}) ~>~ 0. \tag{F}$$
You could definitely use higher order metrics, so that the length element would be $(ds^2)^n$, but that would be non-Riemannian geometry, i.e not GR, plus you have to confront the possibility of $c_n$ light-like characteristics, general Lorentz-invariance failure etc.
Given a vector space $V$, a metric is defined as a map that, given any two elements $v,u\in V$, associates a real positive number $d(v,u) \geq 0$, namely the distance between the two. Since a distance can be derived from a scalar product $\sigma$ by means of $d(v,u) = \sqrt{\sigma(v-u, v-u)}$ assigning a metric is equivalent to assigning a scalar product.
A 2-rank tensor is, by definition, a multilinear map $\tau\colon V\times V\to\mathbb{C}$ and therefore exactly a scalar product. Given $\mathcal{M}$ as space-time manifold with charts $U_i$ and tangent spaces $T_m\mathcal{M}$ in each point $m\in\mathcal{M}$, it is natural to define scalar products in each point as the action of a $(2,0)$-type tensor onto the vectors, evaluated at each point: namely $$ \sigma(X_m, Y_m) = g(m)(X_m, Y_m) $$ induces a positive defined distance in each $T_m\mathcal{M}$ provided $g$ be positive-definite.
This seems to be the most natural choice and although one might use higher rank tensor one would need to evaluate the remaining components onto fixed bases in each $T_m\mathcal{M}$ (to exhaust the remaining entries), which reduces the action to exactly a 2-rank tensor again.