Here's a unifying principle:$\,\,$ Integrating the dyadic product of radial vectors over a sphere will result in an isotropic tensor.
So assume
$$\oint \hat{n}\,\hat{n}\,dA = \beta I$$
for some undetermined scale factor $\beta$.
The scale factor can be determined by contraction
$$\eqalign{
I:I &= \delta_{ij}\,\delta_{ij} = 3 \cr
3\,\beta &= \oint \hat{n}\cdot\hat{n}\,dA \cr &= \oint dA = 4\pi R^2 \cr
\beta &= \frac{4\pi}{3} R^2 \cr
\cr
}$$
Similarly
$$\oint \hat{n}\,\hat{n}\,\hat{n}\,\hat{n}\,dA = \beta\,{\mathbb Y}$$
where ${\mathbb Y}$ is a 4th order isotropic tensor. There are 3 such tensors, so in general ${\mathbb Y}$ is the linear combination
$${\mathbb Y}_{ijkl} = \beta_1\,\delta_{ij}\,\delta_{kl} + \beta_2\,\delta_{ik}\,\delta_{jl} + \beta_3\,\delta_{il}\,\delta_{jk} $$
We obviously need the result which is symmetric over every index, i.e. $$\beta_1=\beta_2=\beta_3=1$$
The contraction $\big\{I:{\mathbb Y}:I\big\}$ yields
$$\eqalign{
\delta_{ij}\,(\delta_{ij}\,\delta_{kl} + \delta_{ik}\,\delta_{jl} + \delta_{il}\,\delta_{jk})\,\delta_{kl}
&= (3\,\delta_{kl} + \delta_{kl} + \delta_{kl})\,\delta_{kl} \cr
&= (5\,\delta_{kl})\,\delta_{kl} \cr
&= 15 \cr
}$$
and the contraction $\big\{I:(\hat{n}\,\hat{n}\,\hat{n}\,\hat{n}):I\big\}$ yields
$$\eqalign{
(\hat{n}\cdot\hat{n})\,(\hat{n}\cdot\hat{n}) &= (1)\,(1) = 1
}$$
So
$$\eqalign{
15\,\beta &= \oint dA = 4\pi R^2 \cr
\beta &= \frac{4\pi}{15} R^2 \cr
}$$
Another approach is to use Gauss' Theorem
$$\eqalign{
\oint \hat{n}\,\hat{n}\,dA
&= \frac{1}{R}\oint\vec{r}\,d\vec{a} \cr
&= \frac{1}{R}\int\nabla\vec{r}\,dv \cr
&= \frac{1}{R}\int\,I\,dv \cr
&= \frac{4\pi}{3} R^2 I \cr
}$$
which matches our previous result.
On the other hand, the isotropic approach tells us immediately that
$$\eqalign{
\oint \hat{n}\,dA = 0 \cr
\oint \hat{n}\,\hat{n}\,\hat{n}\,dA = 0 \cr
}$$
since in the first case, there is no isotropic vector.
And in the second case, there is an isotropic 3rd order tensor (i.e. the Levi-Civita permutation tensor), but it's not symmetric in all indices.
