(Ⅰ) Proof: Since the sequence $\{a_n\}$ has the first term $a_1= \frac{1}{3}$ and common ratio $q= \frac{1}{3}$,
$\therefore a_n=( \frac{1}{3})^{n}$, then $b_n=2\log_{\frac{1}{3}}a_n-1=2\log_{\frac{1}{3}}( \frac{1}{3})^{n}-1=2n-1$.
$\therefore b_{n+1}-b_n=[2(n+1)-1]-(2n-1)=2$.
Thus, the sequence $\{b_n\}$ is an arithmetic sequence with a common difference of $2$;
(Ⅱ) Solution: $c_n=a_n+b_{2n}=( \frac{1}{3})^{n}+(4n-1)$.
$\therefore$ The sum of the first $n$ terms of the sequence $\{c_n\}$, $T_n=c_1+c_2+\ldots+c_n=\left[ \frac{1}{3}+ \frac{1}{3^2}+\ldots+ \frac{1}{3^n}\right]+4(1+2+\ldots+n)-n$
$= \frac{ \frac{1}{3}(1- \frac{1}{3^n})}{1- \frac{1}{3}}+4\cdot \frac{(n+1)n}{2}-n= \frac{1}{2}(1- \frac{1}{3^n})+2n^2+n=2n^2+n+ \frac{1}{2}- \frac{1}{2\cdot 3^n}$.
Thus, the final answer is $T_n = \boxed{2n^2+n+ \frac{1}{2}- \frac{1}{2\cdot 3^n}}$.