Reversing the order and also noting $\binom{n}{k}=\binom{n}{n-k}$ give $$S=(3n+1)\binom{n}{0} + \cdots + 7\binom{n}{n-2}+4\binom{n}{n-1} + 1\binom{n}{n}$$

Adding this to the original relations gives $$2S = (3n+2)\left(\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}\right)=(3n+2)\times 2^n$$

Therefore, $$S={(3n+2)\times 2^{n-1}}$$