I am wondering if there is a simpler way to see this. Computing the cells of the matrix $W(f_1\circ g,...,f_n\circ g)$ row by row, starting at the top, for the first three rows we have with $y=g(x)$ for brevity, the matrix involved$M(f_1\circ g,...,f_n\circ g)$ involved in the Wronskian is $$ M(f_1, \ldots, f_n)(g(x)) = \begin{pmatrix} f_1(y) & f_2(y) & \cdots & f_n(y) \\ f_1'(y)g'(x) & f_2'(y)g'(x) & \cdots & f_n'(y)g'(x) \\ f_1^{(2)}(y)g'^2(x) & f_2^{(2)}(y)g'^2(x) & \cdots & f_n^{(2)}(y)g'^2(x) \\\vdots & \vdots & \vdots & \vdots \end{pmatrix} + \begin{pmatrix} f_1(y) & f_2(y) & \cdots & f_n(y) \\ f_1'(y)g'(x) & f_2'(y)g'(x) & \cdots & f_n'(y)g'(x) \\ f_1'(y)g''(x) & f_2'(y)g''(x) & \cdots & f_n'(y)g''(x) \\\vdots & \vdots & \vdots & \vdots \end{pmatrix}. $$ The second and third rows are linearly dependent (e.g., take the weighted sum of the second row weighted with $g''(x)$ and the third weighted with $-g'(x)$). So for the first three rows, by properties of the determinant we can remove the second term above without changing the value of the determinant. Only the first term survives, which corresponds to the formula you mention, with powers of $g'(x)$ but no higher derivatives of $g$.
For the next rows (4th, 5th etc), say the $k$-th row, only the term in $g'(x)^{k-1}$ survives as all others lead to rows that are linearly dependent with the previous rows (that we have "cleaned" one by one, as for the third row above).
The rest is a combinatorial problem. The second row has $g'(x)$, third row has $g'^2(x)$, fourth row has $g'^3(x)$ etc. As multiplying a row by $C$ multiplies the determinant by $C$, we find a total of $g'(x)^{1+2+...+n-1}$ that can be factored out of the determinant. Since $1+2+...+n-1=n(n-1)/2$, I very much agree with your nice formula!