Defining the determinant of linear transformations as multilinear alternating form

First note that since $T$ is linear, $\mathrm{vol}_T(v_1, \dots, v_n) = \mathrm{vol} (T v_1 , \dots , T v_n )$ is also a multilinear alternating function.

So if you go from $e_1, \dots , e_n$ to a different basis, say, $b_1, \dots b_n$, then you can write the $b_i$ in terms of their coordinates with respect to $e_1, \dots , e_n$:

$$ b_1= \sum_{i=1}^n b_{i 1} \, e_i,\quad b_2= \sum_{i=1}^n b_{i 2} \, e_i,\quad\dots \quad b_n= \sum_{i=1}^n b_{i n} \, e_i. $$

And Leibniz's formula (which is a direct consequence of multinearity and alternating-ness), applied to both $\mathrm{vol}$ and $\mathrm{vol}_T$ tells us:

$$\mathrm{vol}(b_1 , \dots , b_n ) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \: b_{\sigma(1) 1} \cdots b_{\sigma(n) n} \cdot \mathrm{vol}(e_1, \dots , e_n)$$

$$\mathrm{vol} (T b_1 , \dots , T b_n ) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \: b_{\sigma(1) 1} \cdots b_{\sigma(n) n} \cdot \mathrm{vol}(T e_1, \dots , T e_n)$$

So if you take the ratio $\displaystyle \frac{\mathrm{vol} (T b_1 , \dots , T b_n )}{\mathrm{vol} (b_1 , \dots , b_n )} $ everything else cancels out and you're left with $ \displaystyle \frac{\mathrm{vol} (T e_1 , \dots , T e_n )}{\mathrm{vol} (e_1 , \dots , e_n )}$ again.

Of course, for all of this, it is important that $\mathrm{vol} \neq 0$. That's probably an additional restriction your professor put on the $\mathrm{vol}$ function.