Can I use a higher rank tensor field as the metric field? What a metric should necessarily do is it should give me a way to associate a frame invariant number with a given pair of spacetime events. Now, if I use a higher rank tensor field (say, for example, a tensor field of rank 3: $g_{\mu \nu \rho}$) then also I can certainly produce a frame invariant scalar out of a given displacement $\vec{A}$ in this trivial way: $ I := g_{\mu \nu \rho} A^{\mu}A^{\nu}A^{\rho}$. Since the components of the higher rank tensor and that of the displacement vector are going to transform in a covariant and a contravariant manner respectively, the cooked up quantity is certainly a scalar. 
Another crucial property, which I think has more direct physical content than the previous one, is that between two inertial frames, at least one such transformation should exist that leaves at least one metric invariant. i.e., There should exist at least one combination of transformation matrix $\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}$ and metric $g_{\alpha \beta \gamma}$ that satisfies the following equation: 

$\displaystyle\frac{\partial{x^{\alpha}}}{\partial{x^{\mu '}}}\displaystyle\frac{\partial{x^{\beta}}}{\partial{x^{\nu '}}}\displaystyle\frac{\partial{x^{\gamma}}}{\partial{x^{\rho '}}} g_{\alpha \beta \gamma} - \delta_{\mu '}^{\alpha}\delta_{\nu '}^{\beta}\delta_{\rho '}^{\gamma} g_{\alpha \beta \gamma} =0$ 

The last thing I can think of that can put a restriction on the choice of a tensor as a metric is the existence of a possibility of finding a metric compatible symmetric connection field. Following the usual procedure of finding the expression for a metric compatible symmetric connection field, I reached following condition (unlike the case of the usual two rank metric where we get a full-fledged expression) for the connection in the terms of the metric: 

$g_{\nu \rho k_1} \Gamma^{k_1}_{\mu \lambda} - g_{\lambda \mu k_2} \Gamma^{k_2}_{\rho \nu} = \displaystyle\frac{1}{2} (\partial_{\nu}g_{\rho \lambda \mu} + \partial_{\rho}g_{\lambda \mu \nu}  - \partial_{\lambda}g_{\mu \nu \rho}  - \partial_{\mu}g_{\nu \rho \lambda}) $

My question is that if it is possible to satisfy the two highlighted conditions then can we use such a 3 rank (or even higher rank tensors with similarly produced conditions) tensor fields as metric fields? 
PS: This is NOT a proposal for a new home-production theory of gravity (or that of anything for that matter) but rather it is just that I am trying to understand why a two rank tensor is used in General Relativity as the metric. Thank you. 
 A: 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.
A: 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.
A: 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.
--
$^{\dagger}$ We define that the tensor field $g$ is positive definite if 
$$\forall p\in M ~\forall X_p\in T_pM\backslash\{0\}: \quad g_p(\underbrace{X_p,\ldots, X_p}_{r\text{ entries}}) ~>~ 0. \tag{F}$$ 
