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) < 2<sup>2<sup>cn<sup>2</sup></sup></sup> 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 z<sub>0</sub>=0 and z<sub>1</sub>=1, and any z<sub>n</sub> (n≥2) can be obtained from earlier numbers by addition, subtraction, multiplication, division or square root.  The claim follows if |z<sub>n</sub>| < 2<sup>2<sup>O(n<sup>2</sup>)</sup></sup>.

Lemma: The degree of z<sub>n</sub> over ℚ is at most 2<sup>n</sup>.

Proof: Let F<sub>n</sub>=ℚ(z<sub>0</sub>, …, z<sub>n</sub>) be the minimum field containing ℚ∪{z<sub>0</sub>, …, z<sub>n</sub>}.  Then F<sub>0</sub>=ℚ, and F<sub>n</sub> is either equal to F<sub>n−1</sub> or an extension of F<sub>n−1</sub> 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 [F<sub>n</sub>:F<sub>n−1</sub>] is either 1 or 2.  By the degree formula, it holds that [F<sub>n</sub>:ℚ] = [F<sub>n</sub>:F<sub>0</sub>] = [F<sub>n</sub>:F<sub>n−1</sub>][F<sub>n−1</sub>:F<sub>n−2</sub>]…[F<sub>1</sub>:F<sub>0</sub>] ≤ 2<sup>n</sup>.  Therefore, the degree of every element in F<sub>n</sub> over ℚ is also at most 2<sup>n</sup>.  (end of proof of the lemma)

The <em>length</em> 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 2<sup>d<sup>2</sup></sup>((d+1)L)<sup>2d</sup>.  (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 z<sub>n</sub> is at most 2<sup>2<sup>n<sup>2</sup>+n+1</sup></sup> = 2<sup>2<sup>O(n<sup>2</sup>)</sup></sup>.

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, |z<sub>n</sub>| is also at most 2<sup>2<sup>O(n<sup>2</sup>)</sup></sup>.