Let $\{e_1,...,e_n\}$ be standard basis of $\mathbb R^n$, consider the cylinder
$$
L=\{x\in \mathbb R^n : |x-\langle x, e_1\rangle e_1|=1 \}
$$
Obviously, mean curvature of $L$ is $H=n-1$. Assume $\{x_1,..., x_{n-1}\}$ is a local coordinate of $ L$, then , the gradient of $H$ is
$$
\nabla H = g^{ij} \frac{\partial H}{\partial x_i}\frac{\partial}{\partial x_j}=0
$$
On the other hand, the normal vector of $L$ is
$$
\nu=x-\langle x, e_1\rangle e_1
$$
And
$$
\langle x, \nu\rangle =1
$$
So, we have
$$
H=(n-1)\langle x, \nu \rangle
$$
Then,
$$
\nabla \langle x, \nu \rangle =0
$$
Now I want to verify $\nabla \langle x, \nu \rangle =0$.
$$
\nabla \langle x, \nu \rangle =g^{ij}\frac{\partial \langle x, \nu \rangle }{\partial x_i}\frac{\partial}{\partial x_j} \\
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~=g^{ij}
(\langle \nabla_i x, \nu\rangle + \langle x, \nabla _i \nu \rangle)
\frac{\partial}{\partial x_j}
$$
It is equal to show
$$
\langle \nabla_i x, \nu\rangle + \langle x, \nabla _i \nu \rangle =0
$$
Because $\nabla_i x$ is tangent vector, $\langle \nabla_i x, \nu\rangle=0$. Then
$$
\langle x, \nabla _i \nu \rangle = \langle x, -h_{ij}g^{jk}\frac{\partial x}{\partial x_k} \rangle
$$
If the local coordinate is
$$
x_1=e_1, ~~~~\{x_2,...,x_n\} \text{ is the spherical coordinates }
$$
Then, we have
$$
g_{ij}=\delta_{ij} ~~~\text{ and }~~~
h_{ij}=\begin{equation}
\left\{
\begin{aligned}
&1 ~~~~~i=j\ne 1 \\
&0 ~~~~~\text{others}\\
\end{aligned}
\right.
\end{equation}
$$
Then
$$
\langle x, -h_{ij}g^{jk}\frac{\partial x}{\partial x_k} \rangle=0
$$
Since
$$
\langle x, \frac{\partial x}{\partial x_i} \rangle =0 ~~~\text{when $i\ne 1$,}
~~~~~\text{and}~ h_{11}=0
$$
PS: At beginning , I don't know how to calculate $\nabla_i x$ and $\nabla _i \nu$. Then, I know , and add my calculate here.