Does $\mathbb{F}_p((X))$ has only finitely many extension of a given degree? We know that $\mathbb{Q}_p$ has only finitely many extensions of a given degree in its algebraic closure. Is it the same for $\mathbb{F}_p((X))$?
No. There are infinitely many extensions of degree $p$.  Let $K = {\mathbf F}_p((X))$ and $\wp(x) = x^p - x$ (Artin--Schreier operator, which is additive in characteristic $p$). For $c \in K - \wp(K)$, the polynomial $T^p - T - c$ is irreducible in $K[T]$ and its splitting field over $K$ has degree $p$.  For two $c$ and $c'$ in $K - \wp(K)$, the splitting fields of $T^p - T - c$ and $T^p - T - c'$ in an alg. closure of $K$ are equal iff $c = c'$ in $K/\wp(K)$.  This quotient group is infinite. In particular, the reciprocal powers $1/X^n$ for $n$ positive and not divisible by $p$ are distinct in $K/\wp(K)$.  This is discussed in section 1 of http://math.stanford.edu/~conrad/248APage/handouts/weirdfield.pdf.

There are only finitely many extensions of each degree relatively prime to $p$ over ${\mathbf F}_p((X))$ in an algebraic closure, by the same proof as over ${\mathbf Q}_p$. When you deal with extensions of degree divisible by $p$ in characteristic $p$, watch out.
Yes.  The key properties are (1) the field is complete with respect to a nonarchimedean valuation, and (2) the residue field is finite.  With those two properties in hand, then the same proof as for $\mathbb{Q}_p$ goes through.  See (for example) this proof (Theorem 3.19).

EDIT: As others have pointed out, this isn't quite right; the proof fails for extensions of degree divisible by $p$.  See other answers.