I'm not sure how conceptual this is, but I think the explanation is the same as the (intuitively obvious?) fact that most of the area underneath the graph of the function $t^n$ over the unit interval is near $t=1$. In the end it may just be this computational fact, but it is easy enough to see.
Just to flesh things out, consider $\mathbb{S}^n=\{x_1^2+\ldots x_{n+1}^2=1\}$$\mathbb{S}^n=\{x_1^2+\cdots+x_{n+1}^2=1\}$.
By the co-area formula (one can of course appeal to more elementary calculus in the computations below as well) we have that (for $0\leq h\leq 1$)
$$
Vol_n(\mathbb{S}^n\cap \{-h\leq x_1\leq h\})=\int_{-h}^h \int_{\{x_1=t\}\cap \mathbb{S}^n}\frac{1}{|\nabla_{\mathbb{S}^n} x_1|} d\mu dt\\
$$$$
Vol_n(\mathbb{S}^n\cap \{-h\leq x_1\leq h\})=\int_{-h}^h \int_{\{x_1=t\}\cap \mathbb{S}^n}\frac{1}{|\nabla_{\mathbb{S}^n} x_1|} d\mu \, dt.
$$
Observe that $\nabla_{\mathbb{S}^n} x_1$ actually depends only on $t$ (and in particular is independent of $n$).
Indeed,
$$
Vol_n(\mathbb{S}^n\cap \{-h\leq x_1\leq h\})=\int_{-h}^h \frac{1}{\sqrt{1-t^2}}Vol_{n-1}(\{x_1=t\}\cap \mathbb{S}^n) dt
$$$$
Vol_n(\mathbb{S}^n\cap \{-h\leq x_1\leq h\})=\int_{-h}^h \frac{1}{\sqrt{1-t^2}}Vol_{n-1}(\{x_1=t\}\cap \mathbb{S}^n) dt.
$$
Hence,
$$
Vol_n(\mathbb{S}^n\cap \{-h\leq x_1\leq h\})=Vol_{n-1}(\mathbb{S}^{n-1})\int_{-h}^h(1-t^2)^{(n-2)/2} dt
$$$$
Vol_n(\mathbb{S}^n\cap \{-h\leq x_1\leq h\})=Vol_{n-1}(\mathbb{S}^{n-1})\int_{-h}^h(1-t^2)^{(n-2)/2} dt.
$$
Since $0\leq (1-t^2)<1$ when $t\neq 0$, for large $n$ the area under the graph is concentrated near $t=0$, i.e. the equator.
