You can help Mathematica by shifting the d parameter before using Asymptotic:

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

(π^(1/ 2 (-1 + d)) (2 Sqrt[ 2] (-1 + d)^2 (-3 + 2 d) r^4 Hypergeometric2F1[-(1/2), 1/(-4 + 2 d), 1 + 1/(-4 + 2 d), -(1/2) r^(-4 + 2 d)] + Sqrt[2] (-2 + d)  (-1 + d) r^(2 d) Hypergeometric2F1[1/2, 1 + 1/(2 (-2 + d)), 2 + 1/(2 (-2 + d)), -(1/2) r^(-4 + 2 d)] + 2 (-3 + 2 d) r^(  2 + d) (3 -  2 d + (-1 + d) Log[(2 r^d)/( r^d + r^2 Sqrt[2 + r^(-4 + 2 d)])])))/(2 (-1 + d) (-3 + 2 d) r^3 Gamma[(1 + d)/2])

Roman's answer:

roman = 1/Gamma[(1+d)/2] π^(1/2 (-1+d))((Sqrt[2] (-1+d) r^5 (2+r^(-4+2 d))
Hypergeometric2F1[-(1/2),1/(-4+2 d),1+1/(-4+2 d),-(1/2) r^(-4+2 d)])/(2 r^4+r^(2 d))+
(r^(-1+d) (6-4 d+(Sqrt[2] (2-3 d+d^2) r^(2+d) (2+r^(-4+2 d)) 
Hypergeometric2F1[1/2,(3-2 d)/(4-2 d),(7-4 d)/(4-2 d),-(1/2) r^(-4+2 d)])/
((-3+2 d) (2 r^4+r^(2 d)))+2 (-1+d) Log[(2 r^d)/(r^d+r^2 Sqrt[2+r^(-4+2 d)])]))/(2 (-1+d)));

Check:

FullSimplify[asym == roman, d >= 4]

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])
]