What is wrong with this argument?
Let $V$ be a vector space and $g$ an inner product. There exists an orthonormal basis for $V$. That is, in this basis $(g_{ij})=I$. But then given any other basis, the new matrix representation of $g$ is congruent to $I$. Hence the determinant of $g$ in any basis is always 1.
On a manifold you have a metric on the tangent space at each point. But we can always find an orthornomal basis for that tangent space and so $\det g_p=1$. But then this means $\text{vol}_M= \sqrt{\det g}dx^1\wedge\dots\wedge dx^n=dx^1\wedge\dots\wedge dx^n$.
When changing the basis with respect to which a bilinear form is represented, we have$$A'=P^tAP,$$where $A$ and $A'$ are the representing matrices and $P$ is the transition matrix. When one of the matrices represents $g$ with respect to an orthonormal basis, we have$$A'=P^tIP,$$ and thus$$\det A'=\det P^t\cdot\det P=(\det P)^2.$$The last sentence in the question, regarding volume forms, indeed holds.