You are right in saying that $I$ allows one to relate angular velocity and angular momentum in a linear way. It is just not as simple as the momentum and velocity case. An intuition for why things get complicated is that $L = r \times p $ involves a cross product which makes it very sensitive to the choice of a specific set of orthonormal bases(with fixed origin). While $p=mv$ involves a scalar mass that is independent of your choice of coordinates.
To explain inertia tensor, I guess we could start with simpler cases where sufficient symmetry is present (for example a sphere in 3D or a circular pancake in 2d), $L = I(\omega)$ reduces to $L = I\omega$ where $L$ and $\omega$ are vectors and $I$ is just a scalar. An intuition for this reduction is that symmetry makes $I$ resemble $m$ more. As I mentioned earlier, mass is always independent of coordinate choice. but $I$ is only independent of the coordinates that PRESERVE symmetry. Therefore, spheres and circular pancakes are pretty easy to deal with and no inertia tensor is necessary.
But for a general, extended, rigid body in 3d, the lack of symmetry breaks the simple linear relationship. Suppose you have an orthonormal basis, the origin of which is the corner of a cube and the axes line up with the edges of the cube. Basically when the cube is rotated around the z-axis, all the parts of the cube are also instantaneously rotating in the other directions (if you draw a diagram, it would be clear). Therefore $\omega_z$ affects $L_y$ and $L_x$. I don't see an intuitive explanation of the quantitative details.. But this simple cube example shows that $L_x, L_y, L_z$ must each be a linear combination of $\omega_x, \omega_y, \omega_z$. And the mathematical expression that quantifies this must be a matrix.
*A mathematical sidetrack: this matrix itself is not a tensor, but rather a REPRESENTATION of a tensor that maps angular velocity vectors to angular momentum DUAL vectors. In abstract index notation, $L_\alpha = I_{\alpha\beta}\omega^\beta $ You will see a lot of similar notations in E&M, Relativity etc.
Best Answer
This has been answered below, but consider the rotation matrix $\mathrm{R}$ whose columns represent the local $\hat{x}$, $\hat{y}$ and $\hat{z}$ axis in the world coordinates.
$$ \mathrm{R} = \begin{vmatrix} \hat{x} & \hat{y} & \hat{z} \end{vmatrix}\tag{1}$$
This defines the changing inertia tensor from the local inertia tensor $\mathcal{I}_{\rm body}$
$$ \mathcal{I} = \mathrm{R}\, \mathcal{I}_{\rm body} \mathrm{R}^\top \tag{2}$$
Now that rate of change of the rotation matrix is by definition described by the derivative on a rotating frame
$$ \begin{aligned} \frac{\rm d}{{\rm d}t} \boldsymbol{\hat{x}} & = \boldsymbol{\omega} \times \boldsymbol{\hat{x}} \\ \frac{\rm d}{{\rm d}t} \boldsymbol{\hat{y}} & = \boldsymbol{\omega} \times \boldsymbol{\hat{y}} \\ \frac{\rm d}{{\rm d}t} \boldsymbol{\hat{z}} & = \boldsymbol{\omega} \times \boldsymbol{\hat{z}} \\ \end{aligned}$$
or the shorthand rotation $$\dot{\mathrm{R}} = \boldsymbol{\omega} \times \mathrm{R} \tag{3}$$
Now the rate of change of the inertia tensor is the time derivative of (2) and the product rule
$$\require{cancel} \begin{aligned}\dot{\mathcal{I}} & =\mathrm{\dot{R}}\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}+\mathrm{R}\,\cancel{\dot{\mathcal{I}}_{{\rm body}}}\mathrm{R}^{\top}+\mathrm{R}\,\mathcal{I}_{{\rm body}}\dot{\mathrm{R}}^{\top}\\ & =\left(\boldsymbol{\omega}\times\mathrm{R}\right)\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}+\mathrm{R}\,\mathcal{I}_{{\rm body}}\left(\boldsymbol{\omega}\times\mathrm{R}\right)^{\top}\\ & =\boldsymbol{\omega}\times\left(\mathrm{R}\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}\right)-\mathrm{R}\,\mathcal{I}_{{\rm body}}\mathrm{R}^{\top}\left(\boldsymbol{\omega}\times\right)\\ & =\boldsymbol{\omega}\times\mathcal{I}-\mathcal{I}\,\boldsymbol{\omega}\times \end{aligned} \tag{4}$$
with the slight of hand notation $\omega \times$ meaning worry about the cross product later when the tensor is acted upon a vector.
So to derive the equations of motion from the change in angular momentum $\boldsymbol{L} = \mathcal{I} \boldsymbol{\omega}$ you
$$ \tfrac{{\rm d}}{{\rm d}t}\left(\mathcal{I}\boldsymbol{\omega}\right)=\mathcal{I}\dot{\boldsymbol{\omega}}+\left(\boldsymbol{\omega}\times\mathcal{I}-\mathcal{I}\boldsymbol{\omega}\times\right)\boldsymbol{\omega}=\mathcal{I}\dot{\boldsymbol{\omega}}+\boldsymbol{\omega}\times\mathcal{I}\boldsymbol{\omega}-\mathcal{I}\left(\cancel{\boldsymbol{\omega}\times\boldsymbol{\omega}}\right) \tag{5} $$