Well if $\Sigma$ is a submanifold of $R^{n+p}$ and $\{e_i,e_\alpha\}$ is orthonormal frame over $\Sigma$ where the $e_i$'s are tangent and the $e_\alpha$'s are normal to $\Sigma$.
Can anyone prove (with an adequate frame) that $\nabla_{e_i}^{\perp} e_\alpha=0$?
Obs: The result is pretty easy when we have only one normal direction, but in this case there are more.
When we have only one normal direction the picture is somewhat simpler than in the general case.
Introduce $h_i{}^\beta{}_\alpha$ by $$ \nabla^{\perp}_{e_i}e_{\alpha} = h_i{}^\beta{}_\alpha e_\beta $$ (Einstein summation assumed), $i=1,\dots,n$, $\alpha, \beta = n+1,\dots,n+p$, as it is assumed in the question.
Claim. $h_i{}^\beta{}_\alpha = - h_i{}^\alpha{}_\beta$.
Proof. $0 = \nabla_{e_i}\langle e_\alpha , e_\beta \rangle = \langle \nabla^{\perp}_{e_i}e_{\alpha} , e_\beta \rangle + \langle e_\alpha , \nabla^{\perp}_{e_i}e_{\beta} \rangle = h_i{}_\beta{}_\alpha + h_i{}_\alpha{}_\beta$.
Corollary. When $p=1$ (i.e. in the case of hypersurfaces) $$ \nabla_{e_i}^{\perp} e_\alpha=0 $$
Proof. The only skew-symmetric $1\times1$-matrix is $(0)$.