I know that the completely regular spaces are exactly the subspaces of Tychonoff cubes (possibly uncountable powers of $[0,1]$). Out of these, the compact Hausdorff spaces are exactly the closed subspaces of Tychonoff cubes. Is there a way to describe the class of subspaces of Tychonoff cubes that are exactly the class of normal spaces?
Where I've gotten so far:
Sufficient conditions for normality: Second countability, Closedness, Countability of $\lambda$.
Necessary conditions for normality: It isn't this counterexample.
You're really just asking for conditions for a space $X$ to to be $T_4$ (i.e. normal and $T_1$). Any such space embeds as a (thus normal) subspace of $[0,1]^{w(X)}$. E.g. any $F_\sigma$ subset of $[0,1]^I$ is normal (as it is $\sigma$-compact hence Lindelöf and a Lindelöf and regular space is normal. But this is not at all a necessary condition. Even $G_\delta$ subsets need not be normal.
The description is just : the normal subspaces of $[0,1]^I$ are the subspaces that are normal (insert definition here). You cannot expect better. Or you'd just have a characterisation of normality in general.