As suggested in the very nice answer of Prof. Eremenko, I think the key point of this kind of inequality is homogeneity, and it is related to some monotonicity formulae for harmonic functions. We expand briefly below.

If we take the boundary data of $u$ and extend inwards to have homogeneity $\alpha$, i.e. let $v = r^{\alpha}u(x/r)$ (with $r=|x|$) then it is easy to verify
$$|\nabla v(x)|^2 = r^{2\alpha-2}(|\nabla_Tu(x/r)|^2 + \alpha^2u(x/r)^2)$$
where $\nabla_Tu$ is the tangential part of the gradient, giving (in dimension $n$) that
$$(n-2+2\alpha)\int_{B_1} |\nabla v|^2 = \int_{\partial B_1} (|\nabla_Tu|^2 + \alpha^2u^2).$$
Furthermore, since $v$ is a competitor for $u$ we obtain
$$(n-2+2\alpha)\int_{B_1}|\nabla u|^2 \leq \int_{\partial B_1} (|\nabla_Tu|^2 + \alpha^2u^2).$$
Thus, we get a similar inequality to the one you want in $n \geq 3$ by taking $\alpha = 0$. Furthermore, in $\mathbb{R}^2$ we may take $\alpha$ to be the "homogeneity detected at radius $1$",
$$\alpha = \frac{\int_{B_1}|\nabla u|^2}{\int_{\partial B_1}u^2}$$
to get
$$\alpha\int_{B_1}|\nabla u|^2 \leq \int_{\partial B_1} u_{\theta}^2.$$
A very important monotonicity formula of Almgren (very much related to the solution given above by Prof. Eremenko) says that for nonconstant harmonic functions $\alpha \geq 1$ (roughly, the homogeneity detected at radius $r$ increases with $r$ and starts off at least $1$ since $u$ blows up to a plane).

Some nice notes on Almgren's monotonicity are here.

One of its interesting applications is a unique continuation theorem due to Garofalo and Lin, which can be found here.