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,
$$
I = \langle xu^3,yv^3,x^2u^2w-yuvw^3,y^2v^2w-xuvw^2 \rangle.
$$
For the quotient ring $R/I$, there is an injective $k$-algebra homomorphism $S/J\to R/I$, where $S=k[x,y,u,v]$ and $J$ equals $\langle xu^3,yv^3\rangle$.  Moreover, $R/I$ contains a copy of the $S$-module,
$$
S\cdot 1 \oplus S/\langle x^2u^2v^2,y^2u^2v^2\rangle\cdot \overline{w}.
$$
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.  This maximal ideal has height $5$.