Half a year late, but here is a solution. I don't know any good references to your special value, but it can easily be derived using other literature.

From Koekoek and Swarttouw take the Askey-Wilson polynomials
$$
p_n(x;a,b,c,d|q) = \frac{(ab, ac, ad | q)_n}{a^n} {}_4 \phi_{3} (q^{-n}, abcdq^{n-1}, ae^{i\theta}, ae^{-i\theta}; ab, ac, ad; q, q).
$$
It is easily seen that
$$
p_n(\frac{1}{2}(a + a^{-1}); a, b, c, d | q) = a^{-n}(ab,ac,ad;q)_n,
$$
and since the Askey-Wilson polynomials are symmetric in $a$, $b$, $c$ and $d$ you have a similar result for $x = \frac{1}{2}(b + b^{-1})$, $x = \frac{1}{2}(c + c^{-1})$ and $x = \frac{1}{2}(d + d^{-1})$.

Now let's rewrite $H_n(x|q^2)$ in Askey-Wilson polynomials. From exercise 9.10.ii from Basic Hypergeometric Series from George Fasper and Mizan Rahman you have quadratic transformation
$$
H_n(x|q^2) = Q_n(x;q^{1/2},-q^{1/2}|q)
$$
where $Q_n$ are the Al-Salam-Chihara polynomials.

Then following (limit) transformations (http://aw.twi.tudelft.nl/~koekoek/askey/ch4/ch4.html) we rewrite:
$$
Q_n(x;q^{1/2},-q^{1/2}|q) = p_n(x;q^{1/2},-q^{1/2},0,0|q).
$$
And therefore
$$
H_n(\frac{1}{2}(q^{1/2} + q^{-1/2})|q^2) = \frac{(-q;q)_n}{q^{n/2}}.
$$

I hope this helps.