Fields of definition of a variety

No, not in general. Assuming we start with a variety over an algebraically closed field, this would be true if the variety could be defined over its field of moduli, which would then be the unique minimal field of definition. (In particular it is true if the variety has no nontrivial automorphisms.) Note also that it holds trivially for genus zero curves -- there are no moduli! -- and for genus one curves, by the theory of the $j$-invariant.

Therefore the simplest counterexample seems to be a genus $2$ curve over $\mathbb{Q}$. Mestre has shown that the obstruction to such a curve being defined over its field of moduli is a quaternion algebra over $\mathbb{Q}$ -- i.e., the curve can be defined over its field of moduli iff this quaternion algebra is split, i.e., isomorphic to a $2 \times 2$ matrix algebra.

Let $C$ be a genus $2$ curve with field of moduli $\mathbb{Q}$ and nontrivial Mestre obstruction. (Such curves certainly exist. Let me know if you want an explicit example.) Then $C$ may be defined over a quadratic field $K$ iff $K$ splits the quaternion algebra. By the structure theory of quaternion algebras, $C$ can therefore be defined over infinitely many quadratic fields, but not over $\mathbb{Q}$.

Addendum: Here is a nice paper by S. Baba and H. Granath:

http://www.math.kau.se/granath/research/qm.pdf

It deals with the Mestre obstruction on certain genus 2 curves and includes specific examples.

Yes, if varieties are interpreted as subvarieties closed subschemes of base extensions of a fixed ambient variety scheme (e.g., affine space or projective space).

More precisely, suppose that $k \subseteq F$ are fields and the variety $X$ is an $F$-subvariety a closed subscheme of $\mathbf{P}^n_F$. Say for a field $K$ with $k \subseteq K \subseteq F$ that "$X$ is defined over $K$" if $X$ is the base extension of some subvariety of $\mathbf{P}^n_K$. Then $X$ has a minimal field of definition $E$ with $k \subseteq E \subseteq F$, characterized by the property that for any field $K$ with $k \subseteq K \subseteq F$, we have that $X$ is defined over $K$ if and only if $K$ contains $E$.

The same statement holds if $\mathbf{P}^n$ is replaced by any fixed $k$-variety $k$-scheme $Y$.

(Note: this answer does not contradict Pete's. This is just a different interpretation of the question.)

EDIT: As Brian points out, I was indeed assuming that my varieties were closed in the ambient space. The statement about minimal field of definition is not even true for open subschemes in characteristic $p$. For example, if $k=\mathbb{F}_p$ and $F=k(t)$ and $Y=\operatorname{Spec} k[x]$ and $X=\operatorname{Spec} F[x,1/(x-t)]$, then $X$ is the base extension of $\operatorname{Spec} F^{p^n}[x,1/(x^{p^n}-t^{p^n})]$, and hence is definable over $F^{p^n}$ for all $n$, but not over the intersection of all these fields, which is just $k$.

On the other hand, the intersection of any finite number of fields of definition is still a field of definition.

I have generalized to schemes as suggested by Brian.

No, not even for genus $0$ curves, if "$X$ is defined over $K$" means that the variety $X$, initially defined over an extension $F$ of $K$, is to be isomorphic to the base extension of some variety over $K$. (Pete in his answer was implicitly assuming that he was allowed to base extend to an algebraic closure of $F$ before going down to $K$.)

For finite extensions $F \supseteq E$ of $\mathbb{Q}_p$, if $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $E$ if and only if $[F:E]$ is odd (because $\operatorname{Br}(E)[2] \to \operatorname{Br}(F)[2]$ is multiplication by $[F:E]$ from $\frac{1}{2} \mathbb{Z}/\mathbb{Z}$ to itself).

So if $F$ is an $A_4$-extension of $k:=\mathbb{Q}_2$ (such an extension exists), and $K$ and $L$ are the subfields of $F$ fixed by two $3$-cycles in $A_4$, and $X$ is the genus $0$ curve over $F$ corresponding to the non-split quaternion algebra, then $X$ is a base extension of a genus $0$ curve over $K$, and similarly over $L$, but not over $K \cap L = \mathbb{Q}_2$.