I'm not sure how explicit a formula is wanted here, but one can compute differentials if needed by relying on the usual concrete formula in affine algebraic geometry. (This is written down in various textbooks, including section 5.4 of my book on linear algebraic groups where I mostly followed the lecture notes of Borel and Mumford.) For example, since the trace function on $G$ is linear, its differential at a matrix will again be the trace; but calculations for other coefficients of the characteristic polynomial involve more complicated expressions in the matrix coefficients. For general linear groups these can of course be given explicitly.

What Steinberg's map does in the semisimple case comes from his IHES paper on regular elements and is discussed in Chapter 4 of my book on conjugacy classes. His general recipe involves the characters of the $r$ fundamental representations, evaluated at group elements. For general (or special) linear groups, this just amounts to writing down the *significant* coefficients of the characteristic polynomial as noted in the question.

The differential of the Steinberg map can be calculated uniformly at any group element, but since each fiber contains a unique regular class and a unique semisimple class it's usually most interesting to focus on regular elements. This is what Steinberg does in constructing his remarkable cross-section of the fibers.
In this picture the regular semisimple elements are dense, but it's unclear to me why the computation of differentials at such elements is needed. Can you say something about the motivation for the question?