jmc

Since $|a|=1,$ $a\overline{a}=|a{|}^2,$ so $\overline{a}=\frac{1}{a}.$ Similarly, $\overline{b}=\frac{1}{b}$ and $\overline{c}=\frac{1}{c}.$

Also, let $z=a+b+c.$ Then $$\begin{array}{rl}|z{|}^2& =|a+b+c{|}^2\\ & =(a+b+c)(\overline{a+b+c})\\ & =(a+b+c)(\overline{a}+\overline{b}+\overline{c})\\ & =(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\\ & =(a+b+c)\cdot \frac{ab+ac+bc}{abc}\\ & =\frac{a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2+3abc}{abc}.\end{array}$$We have that $$z^3=(a+b+c{)}^3=(a^3+b^3+c^3)+3(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)+6abc,$$so $$\begin{array}{rl}3|z{|}^2& =\frac{3(a^{2}b+a{b}^2+a^{2}c+a{c}^2+b^{2}c+b{c}^2)+3abc}{abc}\\ & =\frac{z^3-(a^3+b^3+c^3)+3abc}{abc}.\end{array}$$From the equation $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1,$ $a^3+b^3+c^3=-abc,$ so $$3|z{|}^2=\frac{z^3+4abc}{abc}=\frac{z^3}{abc}+4.$$Then $$3|z{|}^2-4=\frac{z^3}{abc},$$so $$\left|3|z{|}^2-4\right|=\left|\frac{z^3}{abc}\right|=|z{|}^3.$$Let $r=|z|,$ so $|3r^2-4|=r^3.$ If $3r^2-4<0,$ then $$4-3r^2=r^3.$$This becomes $r^3+3r^2-4=0,$ which factors as $(r-1)(r+2{)}^2=0.$ Since $r$ must be nonnegative, $r=1.$

If $3r^2-4\ge 0,$ then $$3r^2-4=r^3.$$This becomes $r^3-3r^2+4=0,$ which factors as $(r+1)(r-2{)}^2=0.$ Since $r$ must be nonnegtaive, $r=2.$

Finally, we must show that for each of these potential values of $r,$ there exist corresponding complex numbers $a,$ $b,$ and $c.$

If $a=b=1$ and $c=-1,$ then $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1,$ and $$|a+b+c|=|1|=1.$$If $a=1,$ $b=\frac{1+i\sqrt{3}}{2},$ and $c=\frac{1-i\sqrt{3}}{2},$ then $\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}=-1,$ and $$|a+b+c|=|2|=2.$$Therefore, the possible values of $|a+b+c|$ are $\boxed{1,2}.$