Here is another (inductive) proof inspired by Marvis' suggestion below.
It relies on the fact that $\det A \ne0$ iff the columns of $A$ are linearly independent (li.). Let $\Delta_k = \{ (x_1,...,x_k) | x_i \in \mathbb{R}^n, \ x_1,...,x_k \mbox{ are not li.} \}$. Then if we show that $\Delta_n$ has Lebesgue measure zero, it will follow that the set $\{A | \det A = 0 \}$ has Lebesgue measure zero.
In fact, we will show that $m_k \Delta_k = 0$, for $k=1,...,n$, where $m_k$ is the Lebesgue measure on $\mathbb{R}^n \times\cdots \times \mathbb{R}^n$ ($k$ copies).
First we must show that $\Delta_k$ is measurable. If we let $\phi(x_1,...,x_k) = \min_{\|\alpha\| = 1} \| \sum \alpha_i x_i \|$, then we see that $\phi(x_1,...,x_k) = 0$ iff $\{x_1,...,x_k\}$ are not li. Since $\phi$ is continuous, and $\Delta_k = \phi^{-1} \{0 \}$ we see that $\Delta_k$ is Borel measurable.
It is straightforward to see that $\Delta_1 = \{0\}$, hence $m_1 \Delta_1 = 0$. Now suppose this is true for $\Delta_k$, with $1 \leq k < n$. Let $N = \mathbb{R}^n \times \Delta_k$. By assumption $m_{k+1} N = 0$ (since $m_k \Delta_k = 0$). Also, $N \subset \Delta_{k+1}$.
Consider a point $(x, x_1,...,x_k) \in \Delta_{k+1} \setminus N$ (note the indices on the $x$s). Then $\{x_1,...,x_k\}$ are li., but $\{x, x_1,...,x_k\}$ are not li. This can be true iff $x \in \mathbb{sp} \{x_1,...,x_k\}$, a $k$-dimensional hyperplane in $\mathbb{R}^n$ passing through $0$. Note that $m(\mathbb{sp} \{x_1,...,x_k\}) = 0$. Then using Fubini we have (with a slight abuse of notation) $\begin{eqnarray} m_{k+1} (\Delta_{k+1} \setminus N) &=& \int 1_{\Delta_{k+1} \setminus N} \, d m_{k+1}\\ & = & \int 1_{\mathbb{sp} \{x_1,...,x_k\}}(x) 1_{\Delta_k^C}((x_1,...,x_k)) \, d x \, d(x_1,...,x_k)\\ & = & \int \left( \int 1_{\mathbb{sp} \{x_1,...,x_k\}}(x) \, dx \right) 1_{\Delta_k^C}((x_1,...,x_k)) \, d(x_1,...,x_k)\\ & = & \int m(\mathbb{sp} \{x_1,...,x_k\}) 1_{\Delta_k^C}((x_1,...,x_k)) \, d(x_1,...,x_k) \\ & = & 0 \end{eqnarray}$ Since $m_{k+1} (\Delta_{k+1} \cap N) = m_{k+1} N = 0$, we have $m \Delta_{k+1} = 0$. It follows that $m \Delta_k = 0$ for $k=1,...,n$.
If $\mu$ is any measure on $\mathbb{R}^n\times \cdots \mathbb{R}^n$ that is absolutely continuous with respect to the Lebesgue measure ($m_n$ in this case), then it is clear that $\mu \Delta_n = 0$ as well.
In particular, if $\mu$ can be expressed in terms of a joint probability density function, then $\mu \Delta_n = 0$. I believe this includes the case you intended in the question.
I mentioned this in the comments above, but think it is worth repeating here: http://www1.uwindsor.ca/math/sites/uwindsor.ca.math/files/05-03.pdf.