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The original version (with $n = 2^m$) confused $2^{c m^3} = n^{B \log^2 n}$ with $2^{c m^2} = n^{B \log n}$. With the corrected exponent (as pointed out by Will Sawin) this response now gives a complete answer to the question.

Lubotzky proved that the number of $2$-generator groups of order at most $n$ (or order exactly $n$) is bounded by $n^{A \log n}$ for some constant $A$. The number of $2$ groups of order $2^m$ with $n \ge 2^m > n/2$ is (corrected, see Will Sawin's comment) $n^{B \log^2 n}$ for some explicit constant $B$. So you win! Using Lubotzky's theorem, you still win even if $2$-generator groups are replaced by $d$-generator groups for any fixed $d$.

Here is Lubotzky's paper: https://core.ac.uk/download/pdf/82373534.pdf

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