This is a good and difficult question to give a precise answer. From references in Org Syn of phenyl acetylene (
Organic Syntheses, Coll. Vol. 1, p.438 (1941)), both alpha and beta-halostyrenes have been converted. I presume that the halostyrenes are prepared from the dihalides.
An analogous elimination of a 1,2-dibromo-1-phenylethane gave 2-bromo-1-phenylethene with KOtBu (Can J. Chem, 1977). A similar phase transfer elimination gave the alpha-bromo result (
http://www.springerlink.com/content/wjl07u30033m4040/). In that paper they allude to all mechanisms possible. In the abstract, they suggest it is an E1
cb-like reaction.
From that information, the iodide is the preferred leaving group because of the greater acidity of the benzylic hydrogen.
Why is it difficult to know what the product is and the mechanism? Even though I have considerable experience in organic chemistry, reactions can be constructed that are difficult to predict. This is such a case. In this example, a polar solvent is used and the base concentration is not specified. The difference in mechanism in such an instance can vary simply upon dilution. It would be possible to carry out the reaction is a manner more likely to give an E2 elimination (high concentrations, excess base, heat). However, this example allows the elimination to give a Zaitsev or Hoffmann product, different halogens, etc., so it is a difficult question.
Although I could not determine directly, I infer that beta-bromostyrene has been known for a longer time. I presume this is from the elimination from the dibromide. From that model, I would have expected the chloride to be eliminated. However, the leaving bromide has been replaced with a poorer chloride as leaving group and the group to be retained has been replaced with a better leaving iodide. It would be interesting to read a publication with this example.