Explicit bounds for transfer results in algebraic geometry Assume you have an ideal $I\subseteq\mathbb{Z}[X_1,\ldots,X_n]$ of the polynomial ring in $n$ variables over the integers. For any field $\Bbbk$, I can consider the ideal $I(\Bbbk):=I\otimes_{\mathbb{Z}}\Bbbk\subseteq\Bbbk[X_1,\ldots,X_n]$. It is well-known that $1\notin I(\mathbb C)$ implies that for prime numbers $p\gg 0$, we also have $1\notin I(\overline{\mathbb F}_p)$ where $\overline{\mathbb F}_p$ denotes an algebraic closure of the field with $p$ Elements.
Geometrically, it means that the scheme $Z(I)$ which is defined over any field has $\overline{\mathbb F}_p$-rational points for large enough $p$ if (and actually, only if) it has $\mathbb{C}$-rational points. Even more geometric, if the system of equations in $I$ has a solution over the complex numbers, then it admits solutions in characteristic $p$ for large $p$.
I am interested in explicit bounds on $p_0\in\mathbb N$ such that the above holds for all prime numbers $p>p_0$. For this purpose, let us assume that $I$ is generated by polynomially many polynomials of degree less or equal to $d$, for some fixed $d$. Then $p_0=p_0(n)$ depends on the number of variables involved.
In the 1978 paper Bounds on transfer principles for algebraically closed and complete discretely valued fields by Scott Brown, we get a roughly driple exponential bound:

Theorem If $S$ is a first-order statement in the language with plus, minus, times, and constant symbols $0, 1, 10, 11, 100,\ldots$ whose representation is $m$ characters long, and $p>2^{2^{2^{3m}}}$, then $S$ holds in the theory of algebraically closed fields of characteristic $p$ if and only if it holds in the theory of algebraically closed fields of characteristic zero.

Observe that since we fixed the degree $d$, the number $m$ from that theorem will be a polynomial in $n$, because for a fixed degree $d$, the number of monomials in $n$ variables of degree $d$ grows polynomially and I assumed that $I$ is generated by polynomially many elements. 
Edit. I now realize that I also have to assume that the coefficients of my generators are not too large, so let us assume that as well.
Question. Is there a reference for a substantially better bound? I have reasons to believe that a doubly exponential bound (with possibly some polynomial in $n$ in the exponent) should be possible.
 A: The effective Nullstellensatz combined with the bound on the coefficients of the polynomials you are giving us will give you a proof of a bound. Please do not accept this answer; these are not the droids you are looking for.
Namely, say $I = (f_1, \ldots, f_s) \subset \mathbf{Z}[x_1, \ldots, x_n]$ with $f_i$ of degree $d$. Then if modulo $p$ these polynomials generate the unit ideal (this you can test over the field with $p$ elements or its algebraic closure, does not matter which), then we can write
$$
1 = \sum a_i f_i \bmod p
$$
with $\deg(a_i) \leq B = \max(3, d)^n$. This is due to Kollar. Ok, so we look at the map
$$
\mathbf{Z}[x_1, \ldots, x_n]_{\leq B}^{\oplus s} \longrightarrow
\mathbf{Z}[x_1, \ldots, x_n]_{\leq B + d}/\mathbf{Z},\quad
(a_i) \longmapsto \sum a_if_i \mod \mathbf{Z}
$$
This is given by a huge matrix whose entries are bounded. Moreover, the map is injective. Then an easy argument with determinants shows that the map is injective for all $p$ large enough.
How large? Well, just eyeballing what the argument would give you, I think it would not be more than
$$
C^{s{n + B \choose n}}
$$
where $C$ is the bound for the coefficients of the polynomials. I am very bad at doing bounds, so I may be off by quite a large factor. You can also bound $s$ by ${n + d \choose n}$ but I think you should be able to do much better for $s$.
I think the "right" way to answer this question would be to use heights of subvarieties in affine space and an arithmetic nullstellensatz!
