1. We are given that $\alpha$ satisfies $x^2+x+1=0.$ So, $\alpha^2+\alpha+1=0 \implies \alpha^2=-\alpha-1.$

Also, since $\alpha$ is a root of $x^3-1=0$ but not of $x-1=0$, we have $\alpha^3=1, \qquad \alpha\ne 1.$

2. Now compute $(1+\alpha)^7$.

First observe: $1+\alpha=-\alpha^2$ from $1+\alpha+\alpha^2=0.$ Hence, $(1+\alpha)^7=(-\alpha^2)^7=-\alpha^{14}.$

Since $\alpha^3=1$, reduce the power modulo $3$: $14\equiv 2 \pmod 3.$ Therefore, $\alpha^{14}=\alpha^2.$ So, $(1+\alpha)^7=-\alpha^2.$

3. We want to write this in the form $A+B\alpha+C\alpha^2,$ with $A,B,C\ge 0$.

Using $-\alpha^2=1+\alpha$ (because $\alpha^2=-\alpha-1$), we get $(1+\alpha)^7=1+\alpha.$

Thus, $A=1,\quad B=1,\quad C=0.$ These satisfy $A,B,C\ge 0$.

4. Now calculate: $5(3A-2B-C)=5\bigl(3\cdot 1-2\cdot 1-0\bigr)=5(1)=5.$

Therefore, the required integer is $\boxed{5}.$