Let's try writing out the expansion:
$\displaystyle \sum_{k=0}^n \dbinom{2k+1}{2n+1} 2^{3k} = \dbinom{1}{2n+1} + \dbinom{3}{2n+1}2^3 + \dbinom{5}{2n+1}2^6 + \cdots + \dbinom{2n-1}{2n+1}2^{3n-3} + \dbinom{2n+1}{2n+1}2^{3n}$
Every term equals zero except the last term, which equals $2^{3n}$. It appears you have the wrong problem written down. Is it possible you are trying to prove this:
$\displaystyle \sum_{k=0}^n \dbinom{2n+1}{2k}2^{3k} = \dfrac{\left(1+2\sqrt{2}\right)^{2n+1} + (1-2\sqrt{2})^{2n+1}}{2}$?
If that is what you are trying to prove, let's expand the RHS:
$\displaystyle \begin{align*}\dfrac{\left(1+2\sqrt{2}\right)^{2n+ 1} + (1-2\sqrt{2})^{2n+1}}{2} & = \dfrac{\sum_{k=0}^{2n+1}\dbinom{2n+1}{k}2^{3k/2} + \sum_{k=0}^{2n+1}(-1)^k\dbinom{2n+1}{k}2^{3k/2}}{2} \\ & = \dfrac{\left( \dbinom{2n+1}{0} + \dbinom{2n+1}{1}2^{3/2} + \dbinom{2n+1}{2}2^{3} + \cdots + \dbinom{2n+1}{2n}2^{3n} + \dbinom{2n+1}{2n+1}2^{(6n+3)/2}\right) + \left( \dbinom{2n+1}{0} - \dbinom{2n+1}{1}2^{3/2} + \dbinom{2n+1}{2}2^3 \mp \cdots +\dbinom{2n+1}{2n}2^{3n} - \dbinom{2n+1}{2n+1}2^{(6n+3)/2}\right) }{2} \\ & = \dbinom{2n+1}{0} + \dbinom{2n+1}{2}2^3 + \dbinom{2n+1}{4}2^6 + \cdots + \dbinom{2n+1}{2n}2^{3n} \\ & = \sum_{k=0}^n \dbinom{2n+1}{2k}2^{3k}\end{align*}$