Consider the sphere $\mathbb{S}^{n-1}:= \{x \in \mathbb{R}^n : \|x\|_2=1\}$, and let $A^\epsilon_x:= \{z \in \mathbb{S}^{n-1}:\langle z,x \rangle \ge \epsilon\}$ where $x \in \mathbb{S}^{n-1}$. Note that $A^\epsilon_x$ is slightly less than half of $\mathbb{S}^{n-1}$ (if $\epsilon=0$, it'd be exactly half).
We are only concerned with the area/volume of $A^\epsilon_x$, so the $x$ is arbitrary and I will drop the subscript.
In my reading I came across this claim:
For all $\epsilon \in (0,1/\sqrt{2})$, $$\mathbb{P}(A^\epsilon) \le (1-\epsilon^2)^{n/2}.$$
[The probability distribution is the uniform distribution.]
How do you prove this claim?
Let $x = (1,0,0,0,\ldots)$. You want the fraction of sphere with $x_1 \geq \epsilon$. This means all other coordinates are at most $\sqrt{1 - \epsilon^2}$, by Pythagoras' theorem. The volume is at most the volume of a surrounding rectangular box with dimensions $(1-\epsilon)$ in the $x_1$ direction and $(1-\epsilon^2)^{1/2}$ in the other $n-1$ directions. This box volume upper bound is
$ V \leq (1-\epsilon){(1-\epsilon^2)}^\frac{n-1}{2} \leq (1-\epsilon^2)^{n/2}$
This is almost what you requested except it is the fraction of volume, not the surface area.