jmc

We can write the sum as $$\sum _{a_1=0}^{\infty }\sum _{a_2=0}^{\infty }\cdots \sum _{a_7=0}^{\infty }\frac{a_1+a_2+\cdots +a_7}{3^{a_1+a_2+\cdots +a_7}}=\sum _{a_1=0}^{\infty }\sum _{a_2=0}^{\infty }\cdots \sum _{a_7=0}^{\infty }\left(\frac{a_1}{3^{a_1+a_2+\cdots +a_7}}+\frac{a_2}{3^{a_1+a_2+\cdots +a_7}}+\cdots +\frac{a_7}{3^{a_1+a_2+\cdots +a_7}}\right).$$By symmetry, this collapses to $$7\sum _{a_1=0}^{\infty }\sum _{a_2=0}^{\infty }\cdots \sum _{a_7=0}^{\infty }\frac{a_1}{3^{a_1+a_2+\cdots +a_7}}.$$Then $$\begin{array}{rl}7\sum _{a_1=0}^{\infty }\sum _{a_2=0}^{\infty }\cdots \sum _{a_7=0}^{\infty }\frac{a_1}{3^{a_1+a_2+\cdots +a_7}}& =7\sum _{a_1=0}^{\infty }\sum _{a_2=0}^{\infty }\cdots \sum _{a_7=0}^{\infty }\left(\frac{a_1}{3^{a_1}}\cdot \frac{1}{3^{a_2}}\cdots \frac{1}{3^{a_7}}\right)\\ & =7\left(\sum _{a=0}^{\infty }\frac{a}{3^a}\right){\left(\sum _{a=0}^{\infty }\frac{1}{3^a}\right)}^6.\end{array}$$We have that $$\sum _{a=0}^{\infty }\frac{1}{3^a}=\frac{1}{1-1/3}=\frac{3}{2}.$$Let $$S=\sum _{a=0}^{\infty }\frac{a}{3^a}=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\cdots .$$Then $$3S=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+\cdots .$$Subtracting these equations, we get $$2S=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\cdots =\frac{3}{2},$$so $S=\frac{3}{4}.$

Therefore, the given expression is equal to $$7\cdot \frac{3}{4}\cdot {\left(\frac{3}{2}\right)}^6=\boxed{\frac{15309}{256}}.$$