A great question, and there are already some good answers. One way to look at this is through
"Coordinate Free Physics" (https://faculty.washington.edu/seattle/physics544/2011-lectures/thorne-GR.pdf).
The goal is to represent physical law as relations between geometric objects (that do not depend on, nor require, coordinates). As it turns out, these geometric objects are tensors.
I'll be addressing tensors in 3D euclidean space, $\mathbb{E}^3$, in which case tensors are the things that can be rotated "nicely" under the the rotation group ($SO(3)$).
By "nicely" I mean a set of $N$ objects are Eigen-tensors of rotations about an arbitrary axis ($z$), and they are closed under rotations.
(That is, they transform as an $N$-dimensional representation $SO(3)$, but I will avoid technical stuff like that, and only mention it parenthetically).
Step one is to unlearn two things we all have learned.
Numbers are scalars:
While a scalar can be represented by a number, it doesn't mean it is a number. A scalar is a geometric object that is invariant under rotations, so it's an eigenstate of any rotation with eigenvalue $1$. Take, for instance, the pressure at some point, in SI units:
$$ p = \sigma_{ii} = 101325. $$ where $\sigma_{ii}$ is the Cauchy stress tensor field (at the point). The pressure, $p$, is manifestly a scalar, as it is the trace of a rank-2 tensor. It's represented by $101325\,$Pa. If we rotate coordinates, the Cauchy tensor's components change, but the scalar trace is the same, and its representation as a number is unchanged.
The number 101325 is just a number. It can't be rotated, it doesn't live in $\mathbb{E}^3$. It is not a scalar. (Note: unit scalars transform as the trivial representation of $SO(3)$).
Vectors are arrows, they have a magnitude and direction: That's not wrong, you have 3 basis vectors $(\hat x, \hat y, \hat z)$ that span the space of all possible vectors. Intuition goes a long way here, but when moving to higher rank tensors, ones is likely to get lost: what is a dyad, physically? This is because the standard coordinate unit vectors are not fundamental, they transform as the adjoint representation of $SO(3)$.
Now consider so-called spherical vectors (the fundamental representation os $SO(3)$):
$$ \hat e_0 = \hat z $$
$$ \hat e_{\pm 1} = -(\hat x \mp i\hat y)/\sqrt 2 $$
Each basis vector is an eigenvector of rotation about the $z$ axis:
$$ R_z(\theta)[\hat e_m] = e^{im\theta}\hat e_m $$
with eigenvalue:
$$ \lambda = e^{im\theta} $$
For higher ranks $N$, the so-called natural form tensors are symmetric in all indices, and traceless in all indices, so for $N=3$:
$$ T_{ijk} = T_{jik} $$
$$ T_{ijk} = T_{ikj} $$
$$ T_{ijk} = T_{kji} $$
and
$$ T_{iik} = T_{ijj} = T_{kjk} = 0 $$
In general, there are $2N+1$ in each $N$-dimensional irreducible representation of $SO(3)$. The eigenvalue equation is valid for $m\in[-N, -N+1, \ldots, N-1, N]$.
The astute reader will note that that paragraph describes the spherical harmonics, too.
The natural form rank $N$ tensors are the things that can be rotated nicely: they're closed, irreducible, and are eigentensors for $z$-axis rotations.
Physics-wise, we're used to dipole moments being a vector (rank 1). At rank-2, we have quadrupole moments or inertia tensors. We don't see a lot of rank-3 tensors coming up, except the isotropic Levi-Civitta symbol, $\epsilon_{ijk}$.
At rank-4 is the generalized Hooke's law. Here $k$, the spring constant, is generalized to the elasticity tensor, $c_{ijkl}$, which relates the stress and the strain:
$$ \sigma_{ij} = c_{ijkl}\epsilon_{kl} $$
Final note:
At rank-$N$, there are $3^N$ cartesian tensors, but only $2N+1$ natural from tensors (say, for spherical operators in quantum mechanics) and only $N+1$ for classical real tensors (see: solid-harmonics).
The cartesian tensors are reducible, this is the domain of Schur-Wyle duality and the Robertson Schensted correspondence. The integer partitions of $N$ are made into representations of $Sym(N)$, the group of permutations on $N$ letters. From there, the standard Young tableaux each describe the index permutations that form an irreducible representation, and its symmetry. Through the remarkable hook-length formula, the dimensions of the representations can be calculated.
At rank-2, you get:
$${\bf 3}\otimes{\bf 3}={\bf 5}_S\oplus{\bf 3}_A\oplus{1} $$
The ${\bf 5}_S$ is the natural form symmetric tensor, while ${\bf 3}_A$ is the cross-product, and ${\bf 1}$ is the isotropic trace.
tl;dr: rank-$N$ tensors are $N$-nomial products of $(\hat x, \hat y, \hat z)$ that form an $N^3$ reducible representations of the rotation group. That is then a sum of irreducible representations of dimension $2l +1$ for $l \in [1, \cdots, N] a la spherical harmonics. They are the Eigen-shapes of rotations.
Addendum: One of the answers mention spinors. We've all struggled with an intuitive understanding of spinors...the vector with a flag on top that rotates? idk. I think of a spin 1/2 object as the "geometric thing" that can be rotated nicely (under $SU(2)$), where "nicely" again means "a set of things that is closed under rotations" and "are eigenvectors of rotations about the $z$-axis".