The answer is no (and well-known to people working in the representation theory of algebraic groups in positive characteristic). In fact for $V$ finite dimensional and of dimension $>1$ the two vector spaces are not
isomorphic as $GL(V)$-modules ($GL(V)$ is either considered naively as an abstract group when the field $k$
is infinite or as an algebraic group in the general case) for $n=p$ equal to the characteristic.
Under the assumption that $V$ is finite dimensional we may instead formulate the problem as the impossibility
of having a $GL(V)$-isomorphism $Sym^n(V) \cong Sym^n(V^\ast)^\ast$. Now, we have an injective $GL(V)$-map $V^{(p)} \to Sym^p(V)$ given by $v \mapsto v^p$, where $V^{(p)}=k\bigotimes_kV$, where $k$ acts on the left hand side through the $p$'th power (concretely if we choose a basis for $V$ then the action on $V=k^m$ is given by the group homomorphism $GL_m(k) \to GL_m(k)$ which takes $(a_{ij})$ to $(a^p_{ij})$). As $\dim V > 1$ we have $\dim V^{(p)}=\dim V < \dim Sym^p(V)$ so that the inclusion is proper.
It is easily verified that $V^{(p)}$ is irreducible and it is in fact the unique irreducible submodule of $S^p(V)$. This can be seen by starting with an arbitrary non-zero element $f$ of $S^p(V)$ and then acting on it by suitable linear combinations of the action of elementary matrices of $GL(V)$ until one arrives at a non-zero element of (the image of) $V^{(p)}$. This is more easily understood if one uses the fact that we have an action
of an algebraic and consider the induced action by its Lie algebra. Choosing again a basis of $V$ we have elements $x_i\partial/\partial x_j$ whose action on a monomial are very visible. In this way it is clear that
starting with any monomial of degree $p$ one may apply a sequence of such operators to obtain a non-zero
multiple of a monomial of the form $x_k^p$. This plus some thought shows the statement.
Assume now that we have a $GL(V)$-isomorphism $Sym^p(V) \cong Sym^p(V^\ast)^\ast$. Dualising the inclusion $V^{\ast(p)} \hookrightarrow Sym^p(V^\ast)$ and composing with the isomorphism we got a quotient map $Sym^p(V) \to V^{(p)}$. It is easy to see that in Jordan-Hölder sequence of $Sym^p(V)$ $V^{(p)}$ so that the composite $V^{(p)} \to Sym^p(V) \to V^{(p)}$ must be an isomorphism and hence the inclusion $V^{(p)} \hookrightarrow Sym^p(V)$ is split, contradicting that $V^{(p)}$ is the unique simple submodule.
Let me give an alternative proof of the result in Ron Maimon's answer, that the canonical map $V\to V^{**}$ can consistently, in the absence of the axiom of choice, be surjective. The only technical advantage of my proof is that it uses only the consistency of ZF plus countable choice plus "all sets (in Polish spaces) have the Baire property". That consistency was proved (by Shelah) relative to just ZF, whereas the consistency of "all sets are Lebesgue measurable" needs an inaccessible cardinal. I think my proof is also a bit simpler than the one using measure. (Also, I don't need to mention ultrafilters, which some people might consider an advantage.)
Let $f:V^*\to\mathbb R$ be a linear map, and let me identify $V^*$ with the space of infinite sequences of reals. Topologize $V^*$ with the product topology, where each factor $\mathbb R$ has the usual topology of the reals. That makes $V^*$ a Polish space, so I can use the assumption about Baire category. In particular, if I partition $\mathbb R$ into intervals of length 1, then the inverse images of these intervals under $f$ have the Baire property, and they can't all be meager, by the Baire category theorem. So at least one of them, call it $f^{-1}(I)$, differs by a meager set from a nonempty open set. Inside that nonempty open set, I can find a basic open set of the form `B=$\prod_iU_i$ where, for some $n$, the first $n$ of the factors $U_i$ are intervals of some length $\delta$ and the later factors $U_i$ are $\mathbb R$. For the first $n$ indices $i$, let $U'_i$ be the interval with the same midpoint as $U_i$ but only half the length, and let $B'$ be the product that is like $B$ except using the $U'_i$ instead of the $U_i$ for the first $n$ factors. Consider an arbitrary $z\in V^*$ whose first $n$ components are smaller than $\delta/2$ in absolute value. Then translation by $z$ in $V^*$ maps $B'$ homeomorphically to another subset of $B$. So the two sets $f^{-1}(I)\cap B'$ and $\{x\in B':z+x\in f^{-1}(I)\}$ are both comeager in $B'$ and therefore must intersect. Let $x$ be in their intersection. Both $f(x)$ and $f(z+x)$ are in the interval $I$ of length 1. Subtracting (and remembering that $f$ is linear), we get that $|f(z)|\leq 2$.
Summarizing, we have that $|f(z)|\leq 2$ for all $z\in V^*$ whose first $n$ components are smaller than $\delta/2$ in absolute value. By linearity, if the first $n$ components of $z$ are smaller than $\alpha\delta/2$ in absolute value, for some positive $\alpha$, then $|f(z)|\leq2\alpha$. In particular, if the first $n$ components of $z$ are zero, then so is $f(z)$. That is, the kernel of $f$ includes the subspace $N$ consisting of those $z\in V^*$ whose first $n$ components vanish. So $f$ factors through the quotient $V^*/N$, which is finite-dimensional (in fact, $n$-dimensional). Knowing what linear functionals on finite-dimensional spaces look like, we immediately conclude that $f$ is given by inner product with a member of $V$ (having non-zero components in at most the first $n$ positions).
Best Answer
No, it’s not consistent.
Let $V=k^{(\omega)}$ be the vector space of finite sequences of elements of $k$. Then $V^*$ can be identified with the vector space $k^\omega$ of all sequences, and elements of the image of the natural map $V\to V^{**}$, considered as maps $k^\omega\to k$, are determined by their restriction to $k^{(\omega)}$.
So if $V\to V^{**}$ is an isomorphism, then, taking $W=k^\omega/k^{(\omega)}$, there are no nonzero linear maps $W\to k$, and hence $W^{**}=0$. But $W$ is nonzero.
So the map to the double dual must fail to be an isomorphism either for $V$ or for $W$.