More than 99% of groups of order less than 2000 are of order 1024?

In Algebra: Chapter 0, the author made a remark (footnote on page 82), saying that more than 99% of groups of order less than 2000 are of order 1024.

Is this for real? How can one deduce this result? Is there a nice way or do we just check all finite groups up to isomorphism?

Thanks!

Here is a list of the number of groups of order $n$ for $n=1,\ldots,2015$. If you add up the number of groups of order other than $1024$, you get $423{,}164{,}062$. There are $49{,}487{,}365{,}422$ groups of order $1024$, so you can see the assertion is true. (In fact the percentage is about $99.15\%$.)

As far as I know there is no reasonable way to deduce a priori the number of isomorphism classes of groups of a given order, though I believe that combinatorial group theory has some methods for specific cases. A general rule of thumb is that there are a ton of $2$-groups, and in fact I have heard it said that "almost all finite groups are $2$-groups" (though I cannot cite a reference for this statement).

EDIT: As pointed out in the comments, "almost all finite groups are $2$-groups" is still a conjecture. There is an asymptotic bound on the number of $p$-groups of order $p^n$, however. Denoting by $\mu(p,n)$ the number of groups of order $p^n$, $$\mu(p,n)=p^{\left(\frac{2}{27}+O(n^{-1/3})\right)n^3},$$ which is proven here. This colossal growth along with the results of Besche, Eick & O'Brien seem to be what primarily motivated the conjecture.

This is true. The amount of groups of order at most 2000 (up to isomorphism) was calculated precisely for the first time in 2001 by Besche, Eick and O'Brien. Here is the announcement of their result:

We announce the construction up to isomorphism of the $49 910 529 484$ groups of order at most $2000$.

In table 1 the number of groups of order $1024$ is given, it is $49 487 365 422$. Hence ~99.2% of all groups of order at most $2000$ have order $1024$.