Characteristic polynomial of exterior power

I am not sure what data about $f$ you could use other than the characteristic polynomial. The $l$th coefficient of the characteristic polynomial of $\Lambda^k f$ is the trace of $\Lambda^l (\Lambda^k f)$. You would like to write the trace of $\Lambda^l (\Lambda^k f)$ in terms of the trace of $\Lambda^i f$, $1\leq i \leq n$. The unique way to do this is equivalent to describing $\Lambda^l (\Lambda^k V)$ in the representation ring of $GL(V)$ as a polynomial in the $\Lambda^i(V)$, which generate that ring.

So the formulas you are looking for are exactly the polynomials $P_{m,n}$ in the definition of a $\lambda$-ring: http://en.wikipedia.org/wiki/%CE%9B-ring

There they are expressed in terms of symmetric functions in a way that is pretty simple to state but pretty complicated to do computations with. I don't know any clever combinatorial way to simplify these computations, and it seems like a hard combinatorics problem in general. (If you just wanted to get to the symmetric polynomial picture, then going through representation theory might be pointless, as you could just assume $f$ was diagonalizable and go from there.)

Alternately, you could choose to write the coefficients in terms of the eigenvalues of $f$, and then you would write them as a sum over subsets of size $l$ of the set of subsets of size $k$ of the eigenvalues of $f$. You might find this description more elegant.

Finally you could express the $\lambda$-operations in terms of Adams operations by writing the elementary symmetric polynomials in terms of Newton symmetric polynomials. I think this is the easiest way to do computations and remember them and small $k$ and $l$, and is probably the most mathematically interesting way because it is connected to number theory.