One needs to work with the more complicated object $N(\mathbb{Q})\backslash N(\mathbb{A})$ in order to be able to recover the automorphic form. This is called the Whittaker model for $GL_n$, and it goes as follows. Fix a nontrivial character $\psi$ on $\mathbb{Q}\backslash\mathbb{A}$, and consider the Whittaker function $$ W_{\phi}(g):=\int_{N(\mathbb{Q})\backslash N(\mathbb{A})}\phi(xg) \psi(x)^{-1}dx. $$$$ W_{\phi}(g):=\int_{N(\mathbb{Q})\backslash N(\mathbb{A})}\phi(xg) \psi(x_{1,2}+x_{2,3}+\dots+x_{n-1,n})^{-1}dx. $$ The we have the expansion $$ \phi(g)=\sum_{\gamma\in N_{n-1}(\mathbb{Q})\backslash G_{n-1}(\mathbb{Q})} W_\phi\left(\begin{pmatrix}\gamma&0\\0&1\end{pmatrix}g\right).$$ Cogdell in his lecture notes (in Lectures on automorphic $L$-functions, AMS, 2004) says on page 31: "As I said, the proof is not hard. The difficult thing, if there is one, is in recognizing that this is what one needs. This was recognized independently by Piatetski-Shapiro and Shalika."
EDIT. The OP's edit and Paul Garrett's comment show that my answer is not really sastisfactory. Perhaps the real advantage of using $N(\mathbb{Q})\backslash N(\mathbb{A})$ instead of $U(\mathbb{Q})\backslash U(\mathbb{A})$ lies in the fact that the resulting functions $W_\phi$ transform nicely with respect to all unipotent elements, namely $$ W_\phi(xg)=\psi(x_{1,2}+x_{2,3}+\dots+x_{n-1,n})W_\phi(g)\quad\text{for}\quad x\in N(\mathbb{A}). $$