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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 $k[x,y,u,v].$ Let $I$ be the ideal generated by $3$ elements, $$ I = \langle xu^2,yv^2,x^2u-y^2v\rangle.$$ The congruence class of the element $xyuv$ is annihilated by $\langle x,y,u,v\rangle.$ Thus the maximal ideal $\langle x,y,u,v\rangle$ is an embedded prime of $I$. The height of the maximal ideal is $4$, which is strictly larger than $3$.

Here is the result that is true: for an ideal $I$ generated by $n$ elements in a regular local ring $R$, every embedded prime is contained in a minimal prime of height strictly less than $n.$ In particular, for $n=2,$ there can be an embedded prime only if there is a minimal prime of height $1$. For every regular local ring, for every nonzero ideal $I$ generated by $n$ elements, there exists an element $f$ of $R$ and a nonzero ideal $J$ generated by $n$ elements such that $I$ equals $fJ$ and the ideal $J$ has no minimal primes of height $1$. Thus, every associated prime of $J$ has height $2$. Therefore every associated prime of $I$ has height $1$ or $2$. Thus, $n=3$ is the minimal integer such that there exists an ideal $I$ generated by $n$ elements with an embedded prime of height strictly larger than $n$.

Original example. 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 $4.$

For a regular ring $R$, for an ideal $I$ generated by $n$ elements, it is true that every embedded prime contains a minimal prime of height strictly less than $n$. For instance, for the ideal above, the minimal primes are $$\langle x,y\rangle,\ \langle x,v \rangle,\ \langle y,u \rangle,\ \langle u,v \rangle,$$ and these each have height $2$, which is strictly less than $4$.

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 $k[x,y,u,v].$ Let $I$ be the ideal generated by $3$ elements, $$ I = \langle xu^2,yv^2,x^2u-y^2v\rangle.$$ The congruence class of the element $xyuv$ is annihilated by $\langle x,y,u,v\rangle.$ Thus the maximal ideal $\langle x,y,u,v\rangle$ is an embedded prime of $I$. The height of the maximal ideal is $4$, which is strictly larger than $3$.

Here is the result that is true: for an ideal $I$ generated by $n$ elements in a regular local ring $R$, every embedded prime is contained in a minimal prime of height strictly less than $n.$ For every regular local ring, for every nonzero ideal $I$ generated by $n$ elements, there exists an element $f$ of $R$ and a nonzero ideal $J$ generated by $n$ elements such that $I$ equals $fJ$ and the ideal $J$ has no minimal primes of height $1$. In particular, for $n=2,$ there can be an embedded prime only if there is a minimal prime of height $1$. Since every associated prime of $J$ has height $2$, every associated prime of $I$ has height $1$ or $2$. Thus, $n=3$ is the minimal integer such that there exists an ideal $I$ generated by $n$ elements with an embedded prime of height strictly larger than $n$.

Original example. 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 $4.$

For a regular ring $R$, for an ideal $I$ generated by $n$ elements, it is true that every embedded prime contains a minimal prime of height strictly less than $n$. For instance, for the ideal above, the minimal primes are $$\langle x,y\rangle,\ \langle x,v \rangle,\ \langle y,u \rangle,\ \langle u,v \rangle,$$ and these each have height $2$, which is strictly less than $4$.

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 $4.$

For a regular ring $R$, for an ideal $I$ generated by $n$ elements, it is true that every embedded prime contains a minimal prime of height strictly less than $n$. For instance, for the ideal above, the minimal primes are $$\langle x,y\rangle,\ \langle x,v \rangle,\ \langle y,u \rangle,\ \langle u,v \rangle,$$ and these each have height $2$, which is strictly less than $4$.

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 $k[x,y,u,v].$ Let $I$ be the ideal generated by $3$ elements, $$ I = \langle xu^2,yv^2,x^2u-y^2v\rangle.$$ The congruence class of the element $xyuv$ is annihilated by $\langle x,y,u,v\rangle.$ Thus the maximal ideal $\langle x,y,u,v\rangle$ is an embedded prime of $I$. The height of the maximal ideal is $4$, which is strictly larger than $3$.

Here is the result that is true: for an ideal $I$ generated by $n$ elements in a regular local ring $R$, every embedded prime is contained in a minimal prime of height strictly less than $n.$ In particular, for $n=2,$ there can be an embedded prime only if there is a minimal prime of height $1$. For every regular local ring, for every nonzero ideal $I$ generated by $n$ elements, there exists an element $f$ of $R$ and a nonzero ideal $J$ generated by $n$ elements such that $I$ equals $fJ$ and the ideal $J$ has no minimal primes of height $1$. Thus, every associated prime of $J$ has height $2$. Therefore every associated prime of $I$ has height $1$ or $2$. Thus, $n=3$ is the minimal integer such that there exists an ideal $I$ generated by $n$ elements with an embedded prime of height strictly larger than $n$.

Original example. 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 $4.$

For a regular ring $R$, for an ideal $I$ generated by $n$ elements, it is true that every embedded prime contains a minimal prime of height strictly less than $n$. For instance, for the ideal above, the minimal primes are $$\langle x,y\rangle,\ \langle x,v \rangle,\ \langle y,u \rangle,\ \langle u,v \rangle,$$ and these each have height $2$, which is strictly less than $4$.

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 $4$.

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 $4.$

For a regular ring $R$, for an ideal $I$ generated by $n$ elements, it is true that every embedded prime contains a minimal prime of height strictly less than $n$. For instance, for the ideal above, the minimal primes are $$\langle x,y\rangle,\ \langle x,v \rangle,\ \langle y,u \rangle,\ \langle u,v \rangle,$$ and these each have height $2$, which is strictly less than $4$.