Suppose we have $n$ dice. Then
you get $k^2$ with probability ${n\choose k}/2^n$ and your expectation (ignoring the $\\\$70$ fee for playing) is $2^{-n}\sum k^2{n\choose k}$. We have $k{n\choose k}=n{n-1\choose k-1}$ so this equals $2^{-n}n\sum k{n-1\choose k-1}$. Writing $k=(k-1)+1$ and using the same identity with $n-1,k-1$ in place of $n,k$ we see that this equals $2^{-n}n(n-1)\sum {n-2\choose k-2}+2^{-n}n\sum {n-1\choose k-1}=\frac14n(n-1)+\frac12n=\frac14n(n+1)$.
When $n=15$
your expected winnings are \$60 per game, not enough to compensate for the \$70 you pay to play. For that, you need $n(n+1)\geq280$ which first happens at $n=17$.
Just for fun, here's a smartass combinatorial way to prove the identity I used above:
we need $\sum k^2{n\choose k}=2^{n-2}n(n+1)$. The first term is the number of ways to choose some number (say $k$) of balls from a set of $n$, and then choose one of the $k$ twice. Instead of doing that, suppose we pick one ball (from the full set of $n$) twice, and then choose any subset of the others to fill out our set of $k$. There are $n\cdot2^{n-1}$ ways to do that if we pick the same ball twice. There are $2{n\choose 2}\cdot2^{n-2}$ ways to do it if we pick different balls at the start. Adding these gives the required result.
Perhaps there's a smarter-ass way to do it a bit more briefly.