Table over array indices

I think this gets at the core or your question:

Clear[s, t]
n = 3;
vars = Join[Array[Indexed[s, #] &, n], Array[Indexed[t, #] &, n]];
vals = Tuples[{-1, 1}, 2 n];  (* the possible configs *)
rules = Thread[Rule[vars, #]] & /@ vals;  (* rules linking vars to each config *)
vars /. rules  (* an expression in the vars, at each config *)

Here's an evaluation-leak free way: (i.e. it will work even if s[1] etc. have values outside the Table)

n = 3;
Replace[
  Join[s /@ Hold @@ Range@n, t /@ Hold @@ Range@n],
  v_ :> {v, {-1, 1}},
  1
  ] /.
 Hold[s___] :> Hold@Table[(*remove Hold to evaluate*)
    expr,
    s
    ]
(* Hold[
 Table[expr, {s[1], {-1, 1}}, {s[2], {-1, 1}}, {s[3], {-1, 1}}, {t[
    1], {-1, 1}}, {t[2], {-1, 1}}, {t[3], {-1, 1}}]] *)

As noted in the comment above, you'd need to remove the Hold to let the table evaluate. To see how this works, I show some of the individual steps of the above expression:

(* list of s "variables" *)
s /@ Hold @@ Range@n
(* Hold[s[1], s[2], s[3]] *)

(* combine with t "variables" *)   
Join[s /@ Hold @@ Range@n, t /@ Hold @@ Range@n]
(* Hold[s[1], s[2], s[3], t[1], t[2], t[3]] *)

(* insert the iterator specifications *)
Replace[
 Join[s /@ Hold @@ Range@n, t /@ Hold @@ Range@n],
 v_ :> {v, {-1, 1}},
 1
 ]
(* Hold[{s[1], {-1, 1}}, {s[2], {-1, 1}}, {s[3], {-1, 1}}, {t[
   1], {-1, 1}}, {t[2], {-1, 1}}, {t[3], {-1, 1}}] *)

If you don't care about evaluation leaks, this is enough:

n = 3;
Hold@Table[expr, ##] & @@ (
  {#, {-1, 1}} & /@ Join[s /@ Range@n, t /@ Range@n]
  )
(* Hold[
 Table[expr, {s[1], {-1, 1}}, {s[2], {-1, 1}}, {s[3], {-1, 1}}, {t[
    1], {-1, 1}}, {t[2], {-1, 1}}, {t[3], {-1, 1}}]] *)

Again, you'd need to remove the hold in your actual code.

You could also consider using Array if you can change in what way the variables are used:

n = 3;
Array[f[##] &, ConstantArray[2, n], {-1, 1}]
(* {{{f[-1, -1, -1], f[-1, -1, 1]}, {f[-1, 1, -1], 
   f[-1, 1, 1]}}, {{f[1, -1, -1], f[1, -1, 1]}, {f[1, 1, -1], 
   f[1, 1, 1]}}} *)

Here, the function f simply gets the values of the s[…] and t[…] in sequence.

@Alan have gave an elegant way to solve the problem.

Here we just mention that the original expression:

Table[expr[s1, s2, s3, t1, t2, t3], {s1, {-1, 1}}, {s2, {-1, 1}}, {s3, {-1, 1}}, {t1, {-1,1}}, {t2, {-1, 1}}, {t3, {-1, 1}}]

is equivalent to

Outer[expr, {-1, 1}, {-1, 1}, {-1, 1}, {-1, 1}, {-1, 1}, {-1, 1}]