Answer: 349

Solution:

Write $z = a + bi$ with $a^2 + b^2 = p,$ so $p = 2$ or $p \equiv 1 \pmod 4,$ and the pair $\{|a|, |b|\}$ is then unique. Replacing $z$ by $\pm z,$ $\pm\bar{z},$ $\pm iz,$ $\pm i\bar{z}$ only changes the real and imaginary parts of $z^3$ by signs and swaps, so we may take $a \gt b \gt 0,$ and the two candidate side lengths are $|\operatorname{Re} z^3|$ and $|\operatorname{Im} z^3|.$ Expanding $z^3 = (a^3 - 3ab^2) + (3a^2b - b^3)i$ and factoring, $$\begin{aligned} &\operatorname{Re} z^3 + \operatorname{Im} z^3 \\ &\quad = (a - b)(p + 4ab), \qquad \\ &\operatorname{Re} z^3 - \operatorname{Im} z^3 \\ &\quad = (a + b)(p - 4ab). \end{aligned}$$ The triangle exists exactly when $\bigl||\operatorname{Re}| - |\operatorname{Im}|\bigr| \lt p$ $\lt |\operatorname{Re}| + |\operatorname{Im}|,$ and those two quantities are, in some order, the absolute values above. Since $a \gt b$ forces $(a - b)(p + 4ab) \gt p,$ the whole condition reduces to $(a + b)\,|p - 4ab| \lt p.$

Because $a+b\gt\sqrt{a^2+b^2}=\sqrt p,$ this requires $\Delta=|p-4ab|\lt\sqrt p.$ The bound $p=a^2+b^2\lt1000$ gives $1\le b\lt a\le31.$ For each $b\le21,$ substituting $p=a^2+b^2$ into $$(a^2+b^2-4ab)^2\lt a^2+b^2$$ and checking the integers $b\lt a\le31$ leaves only $$\begin{aligned} (a,b,p,\Delta)&=(4,1,17,1), \\ &\quad (7,2,53,3), \\ &\quad (15,4,241,1), \\ &\quad (18,5,349,11). \end{aligned}$$ (This is also a direct finite sieve for every possible prime representation, rather than a search over an unlisted set of primes.) The corresponding values of $(a+b)\Delta$ are $5,27,19,253,$ respectively, all below $p;$ no other pair even passes the necessary weaker inequality. Thus the largest possible prime is $349.$

Indeed for $z = 18 + 5i$ we get $z^3 = 4482 + 4735i,$ and the lengths $349,$ $4482,$ $4735$ form a valid triangle. The answer is $349.$