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Using Asymptotic with an assumption about d ought to do the job (as in Ulrich's answer):

Asymptotic[f, r->Infinity, Assumptions->d>4]

But it takes too long. On the other hand, it looks like you can help Mathematica by shifting the d variable:

asym = Asymptotic[
    FullSimplify[f /. d -> z + 4], 
    r -> Infinity, 
    Assumptions -> z > 0
];

Unfortunately, the output still depends on r:

FreeQ[asym, r]

False

However, if you repeat using Asymptotic on the above output:

asym2 = Asymptotic[asym, r->Infinity, Assumptions -> z > 0];

The output no longer depends on r:

FreeQ[asym2, r]

True

Shifting back and simplifying yields:

r = FullSimplify[asym2 /. z -> d - 4]

-((2^(1/2 (-3 + 1/(-2 + d))) π^(-1 + d/2) Gamma[1/(2 (-2 + d))] Gamma[-((-1 + d)/(2 (-2 + d)))])/Gamma[(1 + d)/2])

This is equivalent to Roman's answer:

Simplify @ Equal[
    r,
    -((2^(1/2 (-3+1/(-2+d))) π^(-1+d/2)*Gamma[-((-1+d)/(2 (-2+d)))]*Gamma[1/(-4+2 d)])/Gamma[(1+d)/2])
]

True