I believe the assertion is true with the additional presumption that $a^2 + b^2 + c^2 \leq 3$.
Let $\phi(x) = x^3 - (a^2 + b^2 + c^2)x + 2 a b c$. Then $M \geq 0$ iff $\phi(x) = 0$ implies $x \leq 1$. Note that $\phi$ is 3rd degree monic polynomial, and that $x$ is a zero of $\phi$ iff $1-x$ is an eigenvalue of $M$.
Suppose $M\geq 0$. Then $\phi(1)\geq 0$, otherwise for some $x>1$ we would have $\phi(x)=0$. It is easy to see that $\phi(1)\geq 0$ is equivalent to the condition $a^2 + b^2 + c^2 - 2 a b c \leq 1$.
Now suppose $a^2 + b^2 + c^2 - 2 a b c \leq 1$. Then we have $\phi(1)\geq 0$. We need to show that $\phi(x)>0$, whenever $x>1$. Since $\phi$ is a monic, cubic polynomial, it will be strictly increasing after any local minimum. We can find the local minimum by looking at $\phi'(x) = 3x^2-(a^2+b^2+c^2)$. We can see that the local minimum occurs at $\hat x =\sqrt{\frac{a^2+b^2+c^2}{3}}$. By assumption, $\hat x \leq 1$, and since $\phi$ is strictly increasing on $[\hat x, \infty)$, and $\phi(1)\geq 0$, we have $\phi(x)>0$ when $x>1$, which finishes the proof.
Addendum: To show the 'iff' mentioned in the comment below, I just need to show that $M\geq 0$ implies $a^2 + b^2 + c^2 \leq 3$. Suppose this is not true, then following the reasoning above, we have that $\hat x > 1$. Substituting the value for $\hat x$, we get $\phi(\hat x) = 2(a b c - {\hat x}^3)= 2(abc - (\frac{a^2+b^2+c^2}{3})^{\frac{3}{2}})$. If I can show that $\phi(\hat x) \leq 0$ this finishes the proof as it implies that $M$ has a negative eigenvalue. Clearly if $abc\leq 0$, then $\phi(\hat x) \leq 0$, so assume that $abc>0$. To conclude I need a small result which is similar to the result that the geometric mean is no bigger than the arithmetic mean:
Suppose $m,n$ are positive integers, and $x_1,...x_n$ are positive. Then we have $\sqrt[n]{x_1...x_n} \leq \sqrt[m]{\frac{x_1^m+...+x_n^m}{n}}.$ By dividing the $\log$ of the right-hand side by $\frac{1}{m}$ and using concavity of $\log$, we have: $\frac{1}{m} \log \frac{x_1^m+...+x_n^m}{n} \geq \frac{1}{m} \sum_{i=1}^n \frac{1}{n} \log x_i^m = \frac{1}{n} \sum_{i=1}^n x_i = \log \sqrt[n]{x_1...x_n}$
Taking the exponential of both sides produces the required identity. Choosing $n=3$, $m=2$, $x_1=|a|, x_2=|b|, x_3=|c|$ we obtain the required result.