Answer: 349

Small Hint:

Write $z = a + bi$ with $a^2 + b^2 = p;$ up to signs and swapping, the triangle needs $\bigl||\operatorname{Re} z^3| - |\operatorname{Im} z^3|\bigr| \lt p.$

Big Hint:

$\operatorname{Re} z^3 - \operatorname{Im} z^3$ $= (a + b)(a^2 + b^2 - 4ab),$ so $4ab$ must be within $\sqrt{p}$ of $p.$

Solution:

Write $z = a + bi$ with $a^2 + b^2 = p.$ The prime $p = 2$ does qualify: for $z = -1 + i$ we have $z^3 = 2 + 2i,$ but it cannot be the largest answer. Hence consider an odd prime $p,$ so $p \equiv 1 \pmod 4$ and the pair $\{|a|, |b|\}$ is 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 $1\le b\lt a\le31;$ moreover $b \le 21,$ because $b \ge 22$ would force $a \ge 23$ and $p \ge 22^2 + 23^2 \gt 1000.$ Substituting $p=a^2+b^2$ into $$(a^2+b^2-4ab)^2\lt a^2+b^2$$ and checking these bounded integers gives the complete list of possible values $$\begin{aligned} p \in \{&10,17,53,68,130,153, \\ &212,241,272,349,386,520, \\ &565,725,778,905,964\}. \end{aligned}$$ Only four values in this list are prime, at $$\begin{aligned} (a,b,p,\Delta)&=(4,1,17,1), \\ &\quad (7,2,53,3), \\ &\quad (15,4,241,1), \\ &\quad (18,5,349,11). \end{aligned}$$ The corresponding values of $(a+b)\Delta$ are $5,27,19,253,$ respectively, all below $p,$ so all four also pass the full triangle 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 $349.$