Suppose you have a sinusoid that has a whole number of cycles ($k$) in your DFT frame contianing $N$ sample points. It can be parameterized like this:
$$ x[n] = A \cos \left( \left( k\frac{2\pi}{N}\right)n + \phi \right) $$
If you take the DFT of this (FFT is a DFT that is computed efficiently), all the bins will be zero except for bins $k$, and $(N-k)$. With MATLAB, bin $k$ occurs at index $k+1$.
$$ X[k] = \frac{A}{2} e^{i\phi} $$
and
$$ X[N-k] = \frac{A}{2} e^{-i\phi} $$
So, you can see, in the ideal case of a pure tone with a whole number of cycles in the frame, the phase angle of the DFT bin corresponds directly to the phase argument in the signal.
The values from $-\pi$ to $\pi$ are by convention and are measured in radians.
If you don't have a whole number of cycles, you can find my simplified bin value formulas here: https://www.dsprelated.com/showarticle/771.php