Lang's observation is: 
>There is a unique linear functional $f\colon S^n(\operatorname{End}V)\to\mathbb K$ such that for any $A\in\operatorname{End}V$,
$$F(A^n)=\det(A).$$

There was another functional $F'\colon S^n(\operatorname{End}V)\to\mathbb K$ that I constructed in the comments above:
> An element $A_1\cdots A_n\in S^n(\operatorname{End}V)$ acts on $\wedge^nV$ by
$$v_1\wedge\cdots\wedge v_n\mapsto \frac1{n!}\sum_{\sigma\in S_n}A_{\sigma1}v_1\wedge\cdots \wedge A_{\sigma n}v_n.$$
Since $\wedge^nV$ is one-dimensional, this linear map is multiplication by a scalar, which I will call $F'(A_1\cdots A_n)$.

I claim that the two linear functionals match. To check this, it suffices to check that $$F'(A^n)=\det(A).$$
But that's because $A^n$ acts on $\wedge^nV$ by
$$v_1\wedge\cdots\wedge v_n\mapsto \frac1{n!}\sum_{\sigma\in S_n}Av_1\wedge\cdots \wedge Av_n=Av_1\wedge\cdots\wedge Av_n,$$
which is simply multiplication by $\det(A)$ by the definition of determinant.