Such a construction works: here's a reference (title: "Construction of a sheaf from the data on a basis of open sets"). Here (pp. 26) is the one pointed out by @DKS in the comments.
Edit (two years later!): A conceptual way to look at this construction is that the formula
\begin{equation}\tag{1}\mathscr{F}(U)\cong\underset{\overset{\longleftarrow}{B\subset U}}{\lim}(\mathscr{F}(B))\end{equation}
is precisely the usual formula computing right Kan extensions. Let $\mathsf{Open}(X)$ be the category whose
- Objects are the open subsets of $X$;
- Morphisms are inclusions.
Given a basis $\mathcal{B}$ of $X$, write $\mathsf{Open}(\mathcal{B})$ for the full subcategory of $\mathsf{Open}(X)$ spanned by the opens of $X$ in $\mathcal{B}$. Then we have a canonical inclusion
$$
i\colon\mathsf{Open}(\mathcal{B})\hookrightarrow\mathsf{Open}(X),
$$
and the presheaf associated to a presheaf $\mathscr{F}\colon\mathsf{Open}(\mathcal{B})^\mathsf{op}\to\mathcal{A}$ on the basis $\mathcal{B}$ taking values on a category $\mathcal{A}$ is precisely the right Kan extension of $\mathscr{F}$ along $i$:

You can go from this to Equation $(1)$ by computing $\mathsf{Ran}_i(\mathscr{F})$ as the limit in Eq. 6.2.3 of Riehl's book:
,
which in our context takes the form
$$
\mathscr{F}(U)\cong\lim\left((\underline{U}\downarrow i)^\mathsf{op}\twoheadrightarrow\mathsf{Open}(\mathcal{B})^\mathsf{op}\overset{\mathscr{F}}{\longrightarrow}\mathcal{A}\right).
$$
But $(\underline{U}\downarrow i)^\mathsf{op}$ is precisely the category whose objects are inclusions of the form $B\hookrightarrow U$ for $B\in\mathcal{B}$, so projecting from this to $\mathsf{Open}(\mathcal{B})^\mathsf{op}$ and then applying $\mathscr{F}$ recovers Eq. $(1)$.
For sheaves, you first take the right Kan extension of $\mathscr{F}$ on $\mathcal{B}$ as above and then sheafify. If $\mathscr{F}$ is already a sheaf on $\mathcal{B}$, then so is the resulting sheaf on $X$ (Görtz–Wedhorn, Prop. 2.20).