Is the number of sequences with equal 0's and 1's small? I think that is n choose n/2. Though when I try to get a feel of what that number should be I get a much larger number than I expect:
i.e. we have
$$ n C \frac{n}{2} = \frac{n!}{\frac{n}{2}! \frac{n}{2}! }$$
using Stirling's?
Recall:
$$n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right) ^n$$
What I tried:
$$ \frac{n!}{\frac{n}{2}! \frac{n}{2}! } 
\sim 
\frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right) ^n}{ \sqrt{2 \pi \frac{n}{2}} \left( \frac{\frac{n}{2}}{e} \right) ^\frac{n}{2}  \sqrt{2 \pi \frac{n}{2}} \left( \frac{\frac{n}{2}}{e} \right) ^\frac{n}{2}} 
\sim
\frac{\sqrt{2 \pi n} \left( \frac{n}{e} \right) ^n}{ \sqrt{2 \pi \frac{n}{2} 2 \pi \frac{n}{2} }  \frac{\frac{n}{2} }{e  ^\frac{n}{2}} ^\frac{n}{2}  \frac{\frac{n}{2} }{e  ^\frac{n}{2}}  ^\frac{n}{2}}
\sim 
\frac{\sqrt{2 \pi n} \frac{n^n}{e^n} }{ \sqrt{ \pi^2 n^2 }  \frac{n^{n}}{e^n} \frac{1}{2^{n}}} \sim \sqrt{2} \frac{2^n}{\sqrt{\pi} \sqrt{n}}
$$
which I feel has to be wrong. 
Why do I feel that? Because n choose n/2 is the number of sequences that have half zero and the other half 1. This number should be much less than the total number of binary sequences (because it you have half 1's and half 0's then to change it to a sequence that is unbalanced should be easy...just change any bit! which means intuitively that half 1's and half zeros should be really small).
Basically, shouldn't most sequences not be balanced? i.e isn't the number of balanced sequences small?
Say the number of not balanced sequences should be
$$2^n - nC\frac{n}{2}$$
shouldn't the above quantity be super large, like $2^n$?
How do I show that the majority of the sequences are unbalanced?

Your derivation is correct, and does tell you that most sequences are unbalanced.  The expression $\sqrt{2} \frac{2^n}{\sqrt{\pi} \sqrt{n}}$ is much smaller than $2^n$ when $n$ is large, by a factor of about $\sqrt{n}$.  For instance, if $n=1000000$, your estimate says that only about $1/1000$ of all the binary sequences are balanced.  The number of unbalanced sequences is about $$2^n-\sqrt{2} \frac{2^n}{\sqrt{\pi} \sqrt{n}}=2^n\left(1-\frac{\sqrt{2}}{\sqrt{\pi}\sqrt{n}}\right),$$
and when $n$ is large, the second factor on the right-hand side is almost $1$.
