Since the hydroxyl is attached, not on the benzene, but on the aliphatic saturated carbon chain, it should react much similarly to the haloalkane:
http://www.chemguide.co.uk/mechanisms/nucsub/hydroxide.html (http://www.chemguide.co.uk/mechanisms/nucsub/hydroxide.html)
That would mean refluxing the reaction with alcohol and base added (probably equal volume to the reaction fluid) - add the base first of course to neutralise the hydriodic acid. Should only take about half an hour, and the result will be no iodocrap.
This would make it very worth the effort of using simple distillation to use the different boiling points of the ephedrine and meth, as there would be no iodoephedrine left after the sodium hydroxide gives it's OH to the iodoephedrine, and steals the iodine.
Treatment with base cyclizes haloephedrines to the corresponding aziridine: Post 483374 (https://www.thevespiary.org/talk/index.php?topic=8340.msg48337400#msg48337400)
(Rhodium: "Synthesis of the required aziridine", Stimulants)
You need some kind of reductive removal. An atmospherical hydrogenation reduces both haloephedrines and aziridines to meth: Post 483349 (https://www.thevespiary.org/talk/index.php?topic=8340.msg48334900#msg48334900)
(Rhodium: ""Ephedrine Aziridine" Reduction to Meth", Stimulants)
Investigation of the impurities found in methamphetamine synthesized from pseudoephedrine by reduction with hydriodic acid and red phosphorus
K. L. Windahl, et. al.
Forens. Sci. Int. 76, 97-114 (1995) (https://www.thevespiary.org/rhodium/Rhodium/chemistry/meth.hi-p.impurities.html)
(https://www.thevespiary.org/rhodium/Rhodium/chemistry/meth.hi-p.impurities.html)