You can do most of the inegrations analyticlly. As the other comments mention, you have to consider singularities.
f = (x1 x2 x3)/((1 + x1^2)^(3/2) (1 + x3^2)^(3/2) (1 + x2^2 -
7 x2 Cos[x4]) Sqrt[x1^2 + x2^2 - x1 x2 Cos[x5]] Sqrt[
1 + 9 (x1^2 + x2^2 - x1 x2 Cos[x6])]) // Together;
int21 = Integrate[f, {x5, 0, 2 \[Pi]}, {x6, 0, 2 \[Pi]},
Assumptions -> {x1 > 0, x2 > 0, x3 > 0, 0 < x4 < 2 Pi}]
(* (16 x1 x2 x3 EllipticK[-((2 x1 x2)/(
x1^2 - x1 x2 + x2^2))] EllipticK[-((18 x1 x2)/(
1 + 9 (x1^2 - x1 x2 + x2^2)))])/(Sqrt[(x1^2 - x1 x2 + x2^2) (1 +
9 (x1^2 - x1 x2 + x2^2))] ((1 + x1^2) (1 + x3^2))^(
3/2) (1 + x2^2 - 7 x2 Cos[x4])) *)
int22 = Integrate[int21, {x3, 0, Infinity},
Assumptions -> {x1 > 0, x2 > 0, x3 > 0, 0 < x4 < 2 Pi}]
(* (16 x1 x2 EllipticK[-((2 x1 x2)/(x1^2 - x1 x2 + x2^2))] EllipticK[-((
18 x1 x2)/(1 + 9 (x1^2 - x1 x2 + x2^2)))])/((1 + x1^2)^(
3/2) Sqrt[(x1^2 - x1 x2 + x2^2) (1 + 9 (x1^2 - x1 x2 + x2^2))] (1 +
x2^2 - 7 x2 Cos[x4])) *)
Singularities at
ContourPlot[(1 + x2^2 - 7 x2 Cos[x4]) == 0, {x2, 0, 10}, {x4, 0,
2 Pi}]
In this range, integral is zero.
int23 = Integrate[int22, {x4, 0, 2 Pi},
Assumptions -> x1 > 0 && x2 > 0, PrincipalValue -> True]
(* ConditionalExpression[0,
3 Sqrt[5] + 2 x2 >= 7 && 2 x2 <= 7 + 3 Sqrt[5]] *)
{red = Reduce[int23[[2]], x2], red // N}
(* {1/2 (7 - 3 Sqrt[5]) <= x2 <= 1/2 (7 + 3 Sqrt[5]),
0.145898 <= x2 <= 6.8541} *)
int24 = Integrate[int22, {x4, 0, 2 Pi},
Assumptions ->
x1 > 0 && (0 < x2 < 1/2 (7 - 3 Sqrt[5]) ||
1/2 (7 + 3 Sqrt[5]) < x2)]
(* (32 \[Pi] x1 x2 EllipticK[-((2 x1 x2)/(
x1^2 - x1 x2 + x2^2))] EllipticK[-((18 x1 x2)/(
1 + 9 (x1^2 - x1 x2 + x2^2)))])/((1 + x1^2)^(
3/2) Sqrt[(x1^2 - x1 x2 + x2^2) (1 - 47 x2^2 + x2^4) (1 +
9 (x1^2 - x1 x2 + x2^2))]) *)
int25 = NIntegrate[
int24, {x1, 0, Infinity}, {x2, 0, 1/2 (7 - 3 Sqrt[5])},
WorkingPrecision -> 15] +
NIntegrate[
int24, {x1, 0, Infinity}, {x2, 1/2 (7 + 3 Sqrt[5]), Infinity},
WorkingPrecision -> 15]
(* 3.820016280311350 *)
Edit
What i did with int23, is regarding x1 and x2 as parameters and calculating the principal value for x4. This is affirmed by inserting parameters.
ii22[x1_, x2_, x4_] = int22; Manipulate[
Plot[ii22[x1, x2, x4], {x4, 0, 2 Pi}, PlotPoints -> 50,
PlotRange -> 5], {{x1, 2}, 0, 10,
Appearance -> "Labeled"}, {{x2, 1/E}, 0, 10,
Appearance -> "Labeled"}]

int23b = Integrate[int22 /. {x2 -> 1/E}, {x4, 0, 2 Pi},
PrincipalValue -> True, Assumptions -> x1 > 0]
(* 0 *)
Reduce[0 < x4 < 2 Pi && 0 == (1 + x2^2 - 7 x2 Cos[x4]) /.
x2 -> 1/E, x4, Reals]
(* x4 == 2 \[Pi] - ArcCos[(1 + E^2)/(7 E)] ||
x4 == ArcCos[(1 + E^2)/(7 E)] *)
NIntegrate[
int22 /. {x1 -> 2, x2 -> 1/E}, {x4, 0, ArcCos[(1 + E^2)/(7 E)],
2 Pi - ArcCos[(1 + E^2)/(7 E)], 2 Pi}, Method -> "PrincipalValue"]
(* 3.71023*10^-14 *)
I'm no pure mathematician to be able to proof this kind of proceeding. But otherwise you must state, integral is not integrable.
1 + x2^2 - 7 x2 Cos[x4] == 0
inx2
andx4
. I have strong doubts about the convergence of the improper integral under consideration. $\endgroup$1 + x2^2 - 7 x2 Cos[x4] ==(x2 - 1/2 (7 Cos[x4] - Sqrt[41 + 49 Cos[2 x4]]/Sqrt[2])*(x2 - 1/2 (7 Cos[x4] + Sqrt[41 + 49 Cos[2 x4]]/Sqrt[2])
. Wee see non- integrable singularities atx2 ==-1/2 (7 Cos[x4] - Sqrt[41 + 49 Cos[2 x4]]/Sqrt[2]
andx2 == 1/2 (7 Cos[x4] - Sqrt[41 + 49 Cos[2 x4]]/Sqrt[2]
inx2
. Don't hesitate to ask for further explanation in need. $\endgroup$