What is the minimum phase shift for Quadrature Amplitude Modulation (QAM)?
The minimum phase shift in 16-QAM, 64-QAM, 256-QAM etc is 0 (same value regardless of whether you are using degrees or radians!) and occurs when the symbol transition is from one constellation point on some radius vector to another constellation point on the same radius vector.  If we take the constellation points as having coordinates $(i,j)$ where $(i,j) \in \{-3, -1, +1, +3\}$ in 16-QAM, then $(1,1)$ and $(3,3)$ lie on the same radius vector (line of slope 1 through the origin) and so the RF phase remains unchanged in the transition from one constellation point to the other. Obviously, the same example will also work for 64-QAM, 256-QAM, and 1024-QAM.

What is the minimum non-zero phase shift?

Well, for 16-QAM, the constellation points $(1,3)$ and $(3,3)$ are distance 2 apart, and together with the origin, form an obtuse triangle with sides of lengths $2$, $\sqrt{10}$, and $\sqrt{18}$. From the standard formula
$$c^2 = a^2 + b^2 - 2ab\cos C$$ relating the sides $a$, $b$, $c$ of a triangle opposite to angles $A$, $B$ and $C$, we get that
$$2^2 = 10 + 18 - 2\sqrt{180} \cos C$$
giving $\arccos \frac{2}{\sqrt{5}}$ as the minimum nonzero phase shift in 16-QAM.
Similar calculations for 64-QAM, 256-QAM and 1024-QAM as left as an exercise for the reader.