One can write more compactly $$\sum_{\substack{0\le m_1<n\\0\le m_2<m_1\\\vdots\\0\le m_{n-2}<m_{n-3}\\0\le m_{n-1}<m_{n-2}}}m_{n-1}.$$ Personally, I would first define a set $$S=\{(m_1,\dots,m_{n-1})\in\mathbb N^{n-1}:n>m_1>m_2>\dots>m_{n-2}>m_{n-1}\ge0\}$$ and then write $$\sum_{(m_1,\dots,m_{n-1})\in S}m_{n-1}.$$
You have a geometric series with common ratio $r=-4$ and scale factor $a=\frac{1}{3}$. See here also. Then the sum of the first $n$ terms is given by
$$\sum_{k=1}^{n}ar^{k-1}=\frac{a(1-r^{n})}{1-r} \quad (1)$$
for $r\neq 1$. Applying this to the above series we obtain
$$\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=\sum_{k=1}^{n}\frac{(-1)^{k}4^{k}}{3}=\sum_{k=1}^{n}\frac{(-4)^{k}}{3}$$
Note that the general term in $(1)$ is $ar^{k-1}$ and at $k=1$ it gives $a$, but in our case, $k=1$ gives $-\frac{4}{3}$. So we can pull out a factor of $-4$ and we have
$$\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=-4\sum_{k=1}^{n}\frac{1}{3}(-4)^{k-1}=-4\frac{\frac{1}{3}(1-(-4)^{n})}{1-(-4)}=\frac{4}{15}((-4)^{n}-1)$$
Where does the above formula come from?
Consider again your example. Let
$$S=\sum_{k=1}^{n}\frac{(-1)^{k}2^{2k}}{3}=-\frac{2^{2}}{3}+\frac{2^4}{3}-\frac{2^6}{3}+...+\frac{(-1)^{n}2^{2n}}{3}$$
$$=\frac{1}{3}\left(-2^2+2^4-2^6+...+(-1)^n2^{2n}\right)=\frac{1}{3}\left(-4+4^{2}-4^{3}+...+(-1)^{n}4^{n}\right)$$
$$=\frac{4}{3}\left(-1+4-4^{2}+...+(-1)^{n}4^{n-1}\right)$$
Then multiply $S$ by $-4$ to obtain
$$-4S=\frac{4}{3}\left(4-4^{2}+4^{3}+...+(-1)^{n+1}4^{n}\right)$$
Next subtract $-4S$ from $S$ to obtain
$$S-(-4S)=\frac{4}{3}\left(-1+4-4^{2}+...+(-1)^{n}4^{n-1}\right)-\frac{4}{3}\left(4-4^{2}+4^{3}+...+(-1)^{n+1}4^{n}\right)$$
$$=\frac{4}{3}\left(-1+\color{red}{4}-\color{green}{4^{2}}+...+\color{blue}{(-1)^n4^{n-1}}-\color{red}{4}+\color{green}{4^2}+...-\color{blue}{(-1)^{n}4^{n-1}}+(-1)^{n}4^{n}\right)$$
$$=\frac{4}{3}\left((-1)^{n}4^n-1\right)$$
$$5S=\frac{4}{3}\left((-4)^n-1\right)\implies S=\frac{4}{15}\left((-4)^n-1\right)$$
Best Answer
Indeed, you are correct, you can use $\sum_{x \in S} C$ for any set $S$ and constant $C$, and since $C$ does not depend on $x$, this simplifies to $$ \sum_{x \in S} C = C \cdot |S|, $$ for any finite set $S$.