Probably because it has a rather complicated branch point at z=1. It is actually very easy to derive the explicit expansion:
singular[x_] = Csc[Sqrt[3] Pi] (-I Sqrt[3] Pi^2 + Pi (1 + 1/(Sqrt[3] (-1 + x)))
- 2 Sqrt[3] Pi HarmonicNumber[Sqrt[3]] - Sqrt[3] Pi Log[-1 + x]);
finite[x_] = (3 - 2 Sqrt[3]) HypergeometricPFQ[{1, 4, Sqrt[3]},
{2 + Sqrt[3], 4 + Sqrt[3]}, 1];
suppressed[x_] = (x - 1) 16/ 13 (-69 + 40 Sqrt[3])
HypergeometricPFQ[{2, 5, Sqrt[3]}, {3 + Sqrt[3], 5 + Sqrt[3]}, 1]
+ (x - 1) 1/2 Pi Csc[Sqrt[3] Pi] (-10 - 9 Sqrt[3] + 10 I Sqrt[3] Pi
+ 20 Sqrt[3] HarmonicNumber[Sqrt[3]] + 10 Sqrt[3] Log[-1 + x]);
asymp[x_] = singular[x] + finite[x] + suppressed[x] (* + O((x-1)^2 Log[x-1]) *);
Notice the logarithmic function in the singular part. Here is the comparison with the original function:
y = 10^-8;
N[HypergeometricPFQ[{4, 4, 1}, {4 + Sqrt[3], 4 - Sqrt[3]}, 1 + y] - asymp[1 + y], 16]
(* -1.892181059711013*10^(-13) + 3.39982418261034*10^(-13) I *)
Derivation
I derived this result by using the integral representation of a generalized hypergeometric function:
integral = HypergeometricPFQ[{a_, b_, c_}, {d_, e_}, x_] :>
Gamma[e]/(Gamma[c] Gamma[e - c]) z^(c - 1) (1 - z)^(e - c - 1) Hypergeometric2F1[a, b, d, x z];
You can easily check it:
Integrate[ Gamma[e]/(Gamma[c] Gamma[e - c]) z^(c - 1) (1 - z)^(e - c - 1)
Hypergeometric2F1[a, b, d, x z], {z, 0, 1}, Assumptions -> Re[c] < Re[e] && Re[c] > 0]
(* HypergeometricPFQ[{a, b, c}, {d, e}, x] *)
Secondly, I applied a simple transformation rule for the integrand:
hyperTransform = HoldPattern[Hypergeometric2F1[a_, b_, c_, z_]] :>
(Gamma[c] Gamma[c - a - b])/(Gamma[c - a] Gamma[c - b])
Hypergeometric2F1[a, b, a + b - c + 1, 1 - z]
+ (1 - z)^(c - a - b) (Gamma[c] Gamma[a + b - c])/(Gamma[a] Gamma[b])
Hypergeometric2F1[c - a, c - b, c - a - b + 1, 1 - z];
Thus, I obtained the following integrand:
Collect[HypergeometricPFQ[{4, 4, 1}, {4 + Sqrt[3], 4 - Sqrt[3]}, x]
/. integral /. hyperTransform, _Hypergeometric2F1, Composition[PowerExpand, FullSimplify]]
which is the sum of two terms:
-Pi x^(-3 + Sqrt[3]) (1 - z)^(-1 + Sqrt[3]) z^Sqrt[3] (1 - x z)^(-1 - Sqrt[3]) Csc[Sqrt[3] Pi]
- (3 + 2 Sqrt[3]) (1 - z)^(-1 + Sqrt[3]) z^3 Hypergeometric2F1[1, 4, 2 + Sqrt[3], 1 - x z]
For the second part, the series expansion about x=1 commutes with the integration with respect to z, hence you can expand it and then integrate, e.g, the leading term is (I made a substitution z-> 1-z)
Integrate[ -(3 + 2 Sqrt[3]) z^(-1 + Sqrt[3]) (1 - z)^3 Hypergeometric2F1[1, 4, 2 + Sqrt[3], z],
{z, 0, 1}]
(* (3 - 2 Sqrt[3]) HypergeometricPFQ[{1, 4, Sqrt[3]}, {2 + Sqrt[3], 4 + Sqrt[3]}, 1] *)
The first term you need to integrate first and then expand:
Integrate[-Pi x^(-3 + Sqrt[3]) (1 - z)^(-1 + Sqrt[3]) z^Sqrt[3]
(1 - x z)^(-1 - Sqrt[3]) Csc[Sqrt[3] Pi], {z, 0, 1}, Assumptions -> Im[x] != 0]
(* -((4^-Sqrt[3] Pi^(3/2) x^(-3 + Sqrt[3]) Csc[Sqrt[3] Pi] Gamma[Sqrt[3]]
Hypergeometric2F1[1 + Sqrt[3], 1 + Sqrt[3], 1 + 2 Sqrt[3], x])/Gamma[1/2 + Sqrt[3]]) *)
FullSimplify[Normal@Series[%, {x, 1, 0}], Assumptions -> x > 1 && Arg[x] == 0]
Thus, you can find any term of the expansion.