How negative can it be

Idea. Not every symmetric matrix with ones on the diagonal is a correlation matrix: it must be positive semidefinite. The eigenvalues of this matrix are easy to find, and asking them to be non-negative gives the bound.

Step 1: the matrix. $$C=\left(\begin{array}{ccc}1& \rho & \rho \\ \rho & 1& \rho \\ \rho & \rho & 1\end{array}\right)=(1-\rho )I+\rho J.$$

Step 2: eigenvalues. $J$ has eigenvalues $3,0,0$, so $C$ has eigenvalues $1+2\rho$ (eigenvector $(1,1,1)$) and $1-\rho$ (twice).

Step 3: the condition. $C$ is positive semidefinite if and only if $1+2\rho \ge 0$ and $1-\rho \ge 0$, that is $-\frac{1}{2}\le \rho \le 1$. The smallest value is $\rho =-\frac{1}{2}$.

Direct check. $\mathrm{Var}(X_1+X_2+X_3)=3+6\rho$ for unit variances, and a variance cannot be negative. At $\rho =-\frac{1}{2}$ the sum is constant: for example, three variables that always add up to zero, with equal variances.

In general. With $n$ variables the bound is $\rho \ge -\frac{1}{n-1}$. Many assets cannot all be strongly negatively correlated with each other.