I am not sure I understand the analogue correctly, but in the commutative case, one can get to depth zero with 3 generators. That is because any second syzygy of a module of depth at least $1$ is isomorphic to a second syzygy of a 3-generated ideal by a result of Bruns. It is even implemented here (be warned that the statement misses the at least depth $1$ part):
http://www.math.uiuc.edu/Macaulay2/doc/Macaulay2-1.2/share/doc/Macaulay2/Bruns/html/
You can take the N to be second syzygy of $m=(t,y_1,...,y_n)$, so $N$ has depth 3. Produce a three generated ideal $I$ such that $syz^2(I)\cong N$. So $depth I =3-2=1$, and $depth R/I=0$.
I think Theorem 13.4 shows that $dim R/I=0$ implies $I$ is at least $n+1$-generated.