The question itself is more general and relates to all p-adic numbers, but it's really easy to show the principle using 2-adics.
The definition of p-adic norm in most textbooks is not easy to understand, but here it is the way I understand it:
The norm of a p-adic number equals $\frac{1}{p^{n-1}}$ with $n$ being the index (the position) of the rightmost non-zero digit we can find.
So, let's start listing all possible 2-adic numbers written with certain precision:
$$ \begin{matrix} & &...0~~& & & &~~...1 & & \\ &...00 & &...10 & & ...01 & & ...11 &\\ ...000 & ...100 & ...010 & ...110 & & ...001 & ...101 & ...011 & ...111 \\ \end{matrix} $$
It can be continued indefinitely. It is easy to see from this pattern that exactly half of 2-adic integers are units, i.e. their norm is $1$, exactly 25% have the norm $\frac{1}{2}$ and so on.
And even more general conclusion can be made using the definition of the distance on 2-adic set - for any given 2-adic integer the distance from it to 50% of other numbers is $1$, to 25% is $\frac{1}{2}$ and so on.
I undertand that it's not a proof, but is there some truth here? Is it correct to say that $\frac{1}{2}$ of 2-adics, $\frac{2}{3}$ of 3-adics, $\frac{4}{5}$ of 5-adics and so on all have the norm $1$?
Best to do this from group theory. First consider the subgroup $p\Bbb Z_p$ of $\Bbb Z_p$. The cosets are the sets $n+p\Bbb Z_p$, for $0\le n<p$. All the elements $a\in n+p\Bbb Z_p$, for $n\ne0$, have $p$-norm equal to $1$. Since the elements of one coset can be put into very natural one-one correspondence with the elements of another, it’s fair to say that $(p-1)/p$ of the elements of $\Bbb Z_p$ have $p$-norm $1$. Now just multiply what I’ve said, and the supporting argument, by $p$: $(p-1)/p$ of the elements of $p\Bbb Z_p$ have $p$-norm $1/p$, right? And so on.