Proofs that every field has a unique (up to isomorphism) algebraic closure use some form of the axiom of choice. For uniqueness this is provably necessary: there are models of ZF in which $\mathbb{Q}$ has two non-isomorphic algebraic closures. Existence is trickier, because algebraic closures of many familiar fields can be constructed by hand.

For example, we can construct an algebraic closure of $\mathbb{Q}$ (or any number field) as the algebraic numbers inside $\mathbb{C}$. More generally, algebraic closures for any countable field can be constructed explicitly (enumerate the irreducible polynomials, adjoin a root to the first and enumerate the new field, factor the second over the new field and adjoin a root of the lexicographically first factor,...). As a further example, the function field of an algebraic curve over $\mathbb{C}$ embeds in the field of Laurent series (choose a local parameter at a point), so this field has an algebraic closure sitting inside the field of Puiseaux series.

Are there examples of specific, explicit fields for which no algebraic closure can be constructed in ZF?

There is a similar question on math.SE with no answers. I tried offering a bounty, to no avail.

isa side comment in the sense that the OP was very clear that his question is about working in ZF, and the comment doesn't answer his question. $\endgroup$ – Todd Trimble♦ Mar 17 '16 at 17:04