Find elements of array with largest absolute value

You could use SparseArray to do this:

sa = SparseArray[H];
With[{ord = Reverse @ Ordering[Abs @ sa["NonzeroValues"], -5]},
    Thread[{
        sa["NonzeroValues"][[ord]],
        sa["NonzeroPositions"][[ord]]
    }]
]

{{2.67802 + 1.28002 I, {1, 2, 2}}, {-2.16128 + 0.250312 I, {3, 2, 1}}, {-0.907637 + 1.71577 I, {3, 3, 1}}, {-1.31683 - 1.21146 I, {1, 2, 3}}, {1.53659 + 0.686038 I, {1, 2, 1}}}

Let h be the OP's H, to avoid single capital letters:

ClearAll[next, arrayPos];
next[{{q_, r_}, d_}] := {QuotientRemainder[r, Times @@ Rest[d]], Rest[d]};
arrayPos[dims_][idx_] := 
  NestList[step, {{0, idx - 1}, dims}, Length[dims]][[2 ;;, 1, 1]] + 1;

arrayPos[Dimensions@h] /@
   Reverse@OrderingBy[Flatten@h, Abs, -5] //
 Transpose[{Extract[h, #], #}] &
(*
{{2.67802 + 1.28002 I, {1, 2, 2}},
 {-2.16128 + 0.250312 I, {3, 2, 1}},
 {-0.907637 + 1.71577 I, {3, 3, 1}},
 {-1.31683 - 1.21146 I, {1, 2, 3}},
 {1.53659 + 0.686038 I, {1, 2, 1}}}
*)

This is shorter and maybe a bit faster than SparseArray:

Ordering[- Flatten @ Abs @ H, top] //
  Tuples[Range @ Dimensions @ H][[#]] & //
  Thread[{Extract[H, #], #}] &

{
 {2.67802 + 1.28002 I, {1, 2, 2}},
 {-2.16128 + 0.250312 I, {3, 2, 1}},
 {-0.907637 + 1.71577 I, {3, 3, 1}},
 {-1.31683 - 1.21146 I, {1, 2, 3}},
 {1.53659 + 0.686038 I, {1, 2, 1}}
}