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An $n$-cube has $\binom{n}{j}2^j$ faces of dimension $n-j$ so the number of cuts is at least $(\sum_0^n\binom{n}{j}2^j)-1-n$. The adjustments are that you seem to want to exclude the $1$ "cut" for $j=0$ which leaves the $n$-cube intact and to only count once each cut into a pair of parallel hyperplanes. If you work out this sum (first without the adjustment terms) I think that you will recognize an exponential growth rate. That gives $3^n-n-1$ which is essential $v^{\log_2{3}}.$ Indeed threshold functions are relevant. I'm not sure what

I found claims that the number is known about their growth rateof order $\binom{2^n}{n}$ which would be $v^{\log{v}}$, more than polynomial but less than exponential.
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An $n$-cube has $\binom{n}{j}2^j$ faces of dimension $n-j$ so the number of cuts is at least $(\sum_0^n\binom{n}{j}2^j)-1-n$. The adjustments are that you seem to want to exclude the $1$ "cut" for $j=0$ which leaves the $n$-cube intact and to only count once each cut into a pair of parallel hyperplanes. If you work out this sum (first without the adjustment terms) I think that you will recognize an exponential growth rate. That gives $3^n-n-1$ which is essential $v^{\log_2{3}}.$ Indeed threshold functions are relevant. I'm not sure what is known about their growth rate.
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An $n$-cube has $\binom{n}{j}2^j$ faces of dimension $n-j$ so the number of cuts is at least $(\sum_0^n\binom{n}{j}2^j)-1-n$. The adjustments are that you seem to want to exclude the $1$ "cut" for $j=0$ which leaves the $n$-cube intact and to only count once each cut into a pair of parallel hyperplanes. If you work out this sum (first without the adjustment terms) I think that you will recognize an exponential growth rate.