I am just posting my comment as an answer.  For a regular ring $R$ and an ideal $I$ generated by $n$ elements, the embedded primes of $I$ can, indeed, have height strictly larger than $n$.  For instance, let $R$ be $k[x,y,u,v,w]$.  Let $I$ be the ideal generated by $4$ elements,
$$
I = \langle xu^3,yv^3,x^2u^2w-yuvw^2,y^2v^2w-xuvw^2 \rangle.
$$
Denote by $S$ the $k$-subalgebra $k[x,y,u,v]$ of $R$.  The $S$-submodule of $R/I$ generated by $\overline{1}$ and $\overline{w}$ is
$$
\left(S/\langle xu^3,yv^3\rangle\cdot 1 \right) \oplus \left( S/\langle xu^3,yv^3,x^2u^2v^2,y^2u^2v^2\rangle\cdot \overline{w}\right).
$$
Thus, the image of $xyu^2v^2w$ in $R/I$ is nonzero.  Yet the annihilator equals all of $\langle x,y,u,v,w\rangle$.  Thus, the maximal ideal $\langle x,y,u,v,w\rangle$ is an embedded prime of $I$.  This maximal ideal has height $5$, which is strictly greater than $4$.