$\Delta \mathbf n = -2 \mathbf n$ on the Euclidean sphere Let us consider the Euclidean two-sphere, defined by the embedding in the three dimensional Euclidean space as
$$
\mathbf n \cdot \mathbf n = 1\,,
$$
where $\cdot$ denotes the standard scalar product. The metric on the sphere, in some coordinates $x^i$, is expressed as 
$$
\gamma_{ij}=\mathbf e_i \cdot \mathbf e_j\,,
$$
where $\mathbf e_i=\partial_i\mathbf n$. For instance, in the standard spherical coordinates 
$$
\mathbf n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)
$$
and
$$
\gamma_{\theta\theta}=1\,,\qquad \gamma_{\phi\phi}=\sin^2\theta\,,\qquad
\gamma_{\theta\phi}=0\,.
$$
We define the Laplace-Beltrami operator on the sphere by
$
\Delta = \gamma^{ij} D_iD_i
$,
where $\gamma^{ij}$ is its inverse and $D_i$ is the associated Levi-Civita connection.
I would like to prove that
$$\Delta \mathbf n = -2 \mathbf n$$
and that in higher dimensions the same holds with $2$ replaced by the dimension of the sphere. I came to believe that this is true by an explicit check in spherical coordinates in dimensions $3$, $4$ and $6$.
Considering that parallel transport of a given tangent vector $\mathbf v$ defined at the point $x+dx$ to the point $x$ is defined by keeping constant its embedding components and then projecting it on the sphere at the point $x$, we have
$$
\mathbf v_{\parallel}(x+dx,x)=\mathbf v(x+dx)-\mathbf v(x+dx)\cdot \mathbf n (x)\, \mathbf n(x)
$$
hence
$$
D_i\mathbf v\, dx^i = \mathbf v_{\parallel}(x+dx,x)- \mathbf v (x)= (\partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v\, \mathbf n)dx^i
$$
where we have used $\mathbf n \cdot \partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v=0$ and
$$
D_i\mathbf v = \partial_i\mathbf v+\partial_i\mathbf n \cdot \mathbf v\, \mathbf n\,.
$$
Applying this to the basis vectors $\mathbf e_j =\partial_j\mathbf n$ affords
$$
D_i\mathbf e_j = \partial_i \mathbf e_j+\gamma_{ij}\mathbf n\,.
$$
But unfortunataly I am not able to go further.

A way to prove the above is the following (this is probably a special case of the more general answer given by @Ernie060, but I still need to fill in a few details). 
In three-dimensional Euclidean space $\mathbb R^3$, the metric in Cartesian coordinates is $\delta_{IJ}=\mathrm{diag}(1,1,1)$ and in spherical coordinates, defined by $\mathbf x = r\,\mathbf n(x^i)$, reads $g_{rr}=1$ and $g_{ij}=r^2\gamma_{ij}$, with $\gamma_{ij}=\partial_i\mathbf n\cdot\partial_j\mathbf n$.  Comparing the two expressions for the Laplacian in the given coordinate systems, we have
$$
0=\Delta_{\mathbb R^{3}}\mathbf x=\frac{1}{r^2}\partial_r(r^2 \mathbf n)+\frac{1}{r}\Delta_{S^2}\mathbf n
$$
which yields precisely
$$
\Delta_{S^2}\mathbf n = -2\mathbf n\,.
$$
This is actually a consequence of the fact that any second covariant derivative of $\mathbf x$ vanishes (since it vanishes in the Cartesian coordiante frame), hence in particular
$$
0=\nabla_i \nabla_j \mathbf x =r D_i D_j\mathbf n-\Gamma_{ij}^r\mathbf n\,,
$$
but explicit calculation affords $\Gamma_{ij}^r=-r\gamma_{ij}$ and hence
$$
D_i D_j \mathbf n = - \gamma_{ij} \mathbf n\,.
$$
