I have often asked myself similar questions. As far as I recall, your question is wide open.

One can however say some things about such a variety:

1) $V$ must have a rational divisor of degree $1$. [Thus a curve with your property must have genus at least $2$.]

Indeed, it has points over both quadratic and cubic fields.

2) $V$ must have points everywhere locally.

Indeed, for all $p \leq \infty$, there exist number fields $K$ strictly larger than $\mathbb{Q}$ which embed in $\mathbb{Q}_p$.

Let me also note that given any finite field $\mathbb{F}_q$, it is not difficult to construct curves which have points over all proper algebraic extensions of $\mathbb{F}_q$ but not over $\mathbb{F}_q$. As a first exercise in this direction, one should determine the finite list of $q$ such that there exists such a genus $2$ curve over $\mathbb{F}_q$ (I am pretty sure that I once wrote down an example to show that this list is nonempty).

Here is a variant of your question that I have thought about enough to be quite sure that I don't know what to do with it: let $[C]: y^2 = P_4(x)$ be a hyperelliptic quartic curve over $\mathbb{Q}$ which has points everywhere locally but no $\mathbb{Q}$-rational points. Thus $C$ represents an order $2$ element of the Shafarevich-Tate group of its Jacobian elliptic curve. Clearly $C$ has many quadratic points: take any $x \in \mathbb{Q}$ and extract the square root of $P_4(x)$. I'm no analytic number theorist, but I believe it is pretty routine to see that $P_4(\mathbb{Q})$ hits infinitely many square classes in $\mathbb{Q}$, so there are infinitely many distinct quadratic splitting fields. If $C$ failed to have a point over $\mathbb{Q}_p$ for some $p$, we could also build infinitely many quadratic *non*splitting fields. But since $C$ has points everywhere locally, I don't know how to rule out the possibility that $C$ has points over **every** quadratic field $\mathbb{Q}(\sqrt{d})$. Does there actually exist such a curve? It seems reasonable to guess *no*, but who knows? Not me.

Finally, your question reminds me of this paper of Jordan Rizov, in which he proves a somewhat similar -- but weaker -- result. I think his paper is nice: he rather cleverly found something along these lines that one can actually prove.