There is an obvious obstacle: the nonreduced scheme $k[x_1, x_2, \ldots, x_n]/\langle x_1, x_2, \ldots, x_n \rangle^2$ is $0$-dimensional, but can't be embedded in any space of dimension less than $n$. More generally, if there is a point whose Zariski tangent space has dimension $n$, then we need $n$ coordinates to embed the scheme. So, for example, if $A$ is the subring of $k[t]$ generated by the monomials $t^n$, $t^{n+1}$, $t^{n+2}$, ..., then $\mathrm{Spec} \ A$ is a reduced one dimensional scheme which can't be embedded in less than $n$ dimensions.

Replace "dimension" by "maximal dimension of any Zariski tangent space" and I think there should be a result like this.

<hr>

The poster clarifies below that he means smooth varieties. In this case, the answer is yes. If $X$ is a smooth projective variety of dimension $d$ over an infinite field then it can be embedded in dimension $2d+1$. The idea of the proof is to embed in $\mathbb{P}^{N-1}$ and consider the Grassmannian of projections $\mathbb{P}^{N-1} \to \mathbb{P}^{2d+1}$. This has dimension $(2d+2)(N-2d-2)$; one shows that the conditions that the projection is not defined on $X$, identifies two points of $X$, or is not injective somewhere on the Zariski tangent space of $X$ all have lower dimension.