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There's not really a fundamental interpretation of the Lagrangian because the Lagrangian that describes the dynamics of a system isn't unique - more than one Lagrangian can yield the correct equations of motion. For instance, let's say we have a particle of mass $m$ experiencing a gravitational force $mg$ in the $-z$ direction. The Lagrangian for this system of the form $L=T-V$ is
$$L(z,\dot{z})=\frac{1}{2}m\dot{z}^2-mgz$$
Plug this into the Euler-Lagrange equations and you should find that the equation of motion is $\ddot{z}=-g$, as expected. However, the following Lagrangian is just as valid:
$$L'(z,\dot{z})=\frac{1}{2}m\dot{z}^2-mgz+\alpha z\dot{z}$$
for some constant $\alpha$. Go ahead and plug it into the Euler-Lagrange equations, and you'll find that it, too, predicts that $\ddot{z}=-g$. The physical interpretation of this function isn't clear, but it turns out to work just as well.

In general, we can add a [total derivative](https://en.wikipedia.org/wiki/Total_derivative) to a system's Lagrangian and get a new Lagrangian that will still yield the same equations of motion. If we take our original Lagrangian $L(z,\dot{z})$ and any function $G(z,\dot{z},t)$, the new Lagrangian
$$L''(z,\dot{z},t)=L(z,\dot{z})+\frac{\partial G}{\partial t}+\frac{\partial G}{\partial z}\frac{\mathrm{d}z}{\mathrm{d}t}+\frac{\partial G}{\partial \dot{z}}\frac{\mathrm{d}\dot{z}}{\mathrm{d}t}$$
will give us the same equation of motion.

In fact, a valid Lagrangian in classical mechanics doesn't even need to have the units of energy. There are some absurd functions you can construct that are still valid Lagrangians, [such as](https://physics.stackexchange.com/a/17407/56299)
$$L=\frac{1}{3}T^2+2TV-V^2$$
All of this should hopefully convince you that it's not really possible to ascribe a physical meaning to a Lagrangian, even in classical mechanics. The quantity $T-V$ does have a (trivial) meaning - the difference between the kinetic and potential energies - but it is only one of many possible valid Lagrangians.