Edited answer (see below for my original, less useful reply): Since your polynomial p(z) of degree 2d is palindromic, rewrite it as z^d q(z+1/z) for some polynomial q(x) of degree d. Then p(z) has 2d roots on the circle if and only if q(x) has d real roots in the interval [-2,2]. (Equivalently, q(x-2) should have d nonnegative real roots, while q(x+2) should have no positive roots.) Now you can try to extract information using e.g. Sturm sequences to try to count real roots in that interval.
I had previously posted: Since the real line parameterizes the circle (minus a point), you can transform your problem into counting the real zeros of an associated real polynomial. (See p. 182 of Rodriguez Villegas's book "Experimental Number Theory" for details; this is viewable in Google Books.)
There certainly have to exist necessary & sufficient criteria for all the zeros of a real polynomial to be real, involving rather complicated inequalities in the coefficients of the polynomial, generalizing the condition b^2-4ac >= 0 for quadratics; or for a specific polynomial you can try a real-root-counting algorithm (see Sturm's theorem on Wikipedia).
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