This is not a complete answer to the question, but here is a proof that D(n) is at most doubly-exponential with quadratic exponent, i.e., D(n) < 22cn2 for some positive constant c>0 for sufficiently large n.
As Terry Tao pointed out, this problem can be rephrased in the algebraic form. We have z0=0 and z1=1, and any zn (n≥2) can be obtained from earlier numbers by addition, subtraction, multiplication, division or square root. The claim follows if |zn| < 22O(n2).
Lemma: The degree of zn over ℚ is at most 2n.
Proof: Let Fn=ℚ(z0, …, zn) be the minimum field containing ℚ∪{z0, …, zn}. Then F0=ℚ, and Fn is either equal to Fn−1 or an extension of Fn−1 obtained by adjoining a square root. Since adjoining a square root of a non-square element gives an extension of degree 2, the extension degree [Fn:Fn−1] is either 1 or 2. By the degree formula, it holds that [Fn:ℚ] = [Fn:F0] = [Fn:Fn−1][Fn−1:Fn−2]…[F1:F0] ≤ 2n. Therefore, the degree of every element in Fn over ℚ is also at most 2n. (end of proof of the lemma)
The length of an algebraic number α is the sum of the absolute values of the coefficients of the integer-coefficient minimum polynomial of α. If algebraic numbers α and β have degree at most d and length at most L, then α+β, α−β, αβ, α/β and √α have length at most 2d2((d+1)L)2d. (I obtained this by constructing the minimum polynomials of α+β etc. by using resultants and dirty calculation. I do not claim that this bound is close to optimal.) By combining this with the earlier bound on the degree and using the mathematical induction, we obtain that the length of zn is at most 22n2+n+1 = 22O(n2).
Since the length is an upper bound on the height and the absolute value of a root of a polynomial is at most 1 plus the height of the polynomial, |zn| is also at most 22O(n2).