The answer is no except for the trivial case $\mathbb F =\mathbb F_2$ (in which $V=0$)
Indeed, assume $\mathbb F$ is such a field and take $x\in \mathbb F^*$. In the structure of $V$, we have $2\cdot x = 0$, which translates to $x^2= 1$ in $\mathbb F$ (for each coordinate).
This implies that any element of $\mathbb F$ is a root of $X^2-1$ which has at most to roots : $|\mathbb F| \leq 2$
As suggested in the comments, let us try to see what happens if we have two fields a field $K$, a field $\mathbb F$ and we want to see when $K^*$ has a $\mathbb F$ vector space structure. Suppose $K$ has characteristic $\neq 2$. Then $(-1)^2 = 1$ implies that either $K^* = 0$ as a vector space (and so $K=\mathbb F_2$, conversely, this can be given a vector space structure over any field) or $2=0$ in $\mathbb F$, so $\mathbb F$ has characteristic $2$. But then $x^2=1$ for all $x\in K^*$ which implies that $K^*$ has at most two elements, in particular $K=\mathbb F_2$ or $\mathbb F_3$. Conversely, $\mathbb F_3^*$ is a $\mathbb F_2$-vector space.
Now if $K$ has characteristic $2$: as Jyrki Lahtonen points out, at least all the Mersenne primes give fields over which it works. I haven't figured out the whole thing yet.
If $K$ has characteristic $2$ (and is not $\mathbb F_2$) then $\mathbb F$ can't, therefore $2$ is invertible : in particular $K$ must be a perfect field as every element has a square root.
If $\mathbb F$ is of positive characteristic, then in particular $p=0$ for some $p$ so that $K$ must be finite, and in particular $K^*$ is cyclic : it must be $1$-dimensional over the prime subfield of $\mathbb F$ which must therefore equal $\mathbb F$, say it is $\mathbb F_p$ and then $p$ is a Mersenne prime, and conversely any Mersenne prime yields an example.
It remains to see what happens if $\mathbb F$ has characteristic $0$. In this case any element of $K^*$ is uniquely divisible. It follows that $K$ can't contain any nontrivial finite field : if $x$ is algebraic over $\mathbb F_2$, then it has exactly $3$ cube roots, all in the same finite subfield, instead of just the one.