Is unirationality decidable? When the answer to the Lüroth problem is affirmative, the genus (for curves) or Castelnuovo's criteria (for separably unirational surfaces) give computable invariants which decide if a given variety is unirational.
I am interested in the simplest cases where non-rational examples are known: namely surfaces over $\overline{\mathbb{F}_p}$ and 3-folds over $\mathbb{C}$.
Is it known whether there exists an algorithm to determine if a given variety of the above type is unirational?
A naive hope would be to produce a bound on the degree of a rational map $\mathbb{P}^n \to X$ however this is clearly impossible since rational maps $\mathbb{P}^n \to \mathbb{P}^n$ have unbounded degree.
The next best thing would be to put a computable upper bound on the minimal degree of such a rational map. Are such bounds known for the two cases I outlined above? Are such bounds sufficient, in principle, to provide an algorithm?
 A: This is a widely studied problem, and I think with the current technology it looks unlikely this will be answered. Let me stick to characteristic $0$ for simplicity; the story gets much richer in positive characteristic (even if you add the adjective 'separably' everywhere).
Unirational varieties are rationally connected, which in turn implies $H^0(X,(\Omega_X^1)^{\otimes m}) = 0$ for all $m > 0$. Mumford conjectured that this vanishing conversely implies that $X$ is rationally connected, but I think little is known about this conjecture. At least if you believe Mumford (and maybe with a bound on which $m$ you have to try, analogous to Castelnuovo's criterion), then rational connectedness should be somewhat decidable.
However, a well-known open problem is whether every rationally connected variety is unirational (I think people expect this to be false ― I certainly do). The reason we can't answer this is that we don't have any obstructions that can distinguish between the two. Therefore, I think it's highly unlikely that we can give a characterisation that decides whether a variety is unirational.
I'm not sure what the status is on unirationality in families, but (stable) rationality is not a deformation invariant of smooth projective varieties by Hassett–Pirutka–Tschinkel. So discrete [i.e. locally constant in families] invariants like cohomology vanishing are not enough to detect (stable) rationality, and very likely the same is true for unirationality.

As for surfaces over $\bar{\mathbf F}_p$, some K3 surfaces are unirational, but most are not, so once again it is not a deformation invariant. Shioda conjectured that if $X$ is a surface over an algebraically closed field $k$ of characteristic $p > 0$ with $\pi_1^{\operatorname{\acute et}}(X) = 0$, then $X$ is unirational if and only if $H^2_{\operatorname{cris}}(X/K)$ is supersingular, i.e. all slopes of Frobenius are $1$. (This is clearly a necessary condition, since $H^2_{\operatorname{cris}}(X/K) \hookrightarrow H^2_{\operatorname{cris}}(Y/K)$ is injective and preserves Frobenius actions if $Y \twoheadrightarrow X$ is a dominant map of smooth projective varieties; see e.g. Prop. 1.2.4 in Kleiman's paper Algebraic cycles and the Weil conjecutres in Dix Exposés.)
But this doesn't say anything about what happens if $\pi_1^{\operatorname{\acute et}}(X) \neq 0$, although it has to be finite and its order indivisible by $p$ (see e.g. this note by Chambert-Loir). So you're first supposed to compute the universal cover and then compute slopes ― I'm not sure if there is a more direct way to do this. Assuming Shioda's conjecture this gives a criterion.
