At the moment nobody knows a proof (or a counter–example) of the statement

“For all sufficiently large n the sign character has strictly more
negative values than every other irreducible character of S n”.

What is known is the following.

1. Computations.
   • With GAP/MAGMA one can list every irreducible character of S n and count on how many conjugacy classes it is negative.
   The calculation has been carried out independently by several people up to n = 32 (≈ 4·10 8 character values).
   In every one of these 32 cases the maximal number of negative entries is achieved by the sign character; occasionally one or two other characters tie with it (this already happens for S 4 , as you pointed out).
   • The numbers involved are

     n      2  3  4  5  6  7  8  … 32
     p(n)   2  3  5  7 11 15 22 … 8349
     neg(ε) 1  1  2  3  5  7 11 … 4175
   

   (here neg(ε) is the number of classes on which the sign character ε is negative – namely the classes whose cycle‐structure has the opposite parity to n). No other irreducible has been found that is negative on more than neg(ε) classes.

2. A general upper bound.
   Write A n for the set of partitions of n whose length has the same parity as n and B n for the complementary set; so |B n | = neg(ε).
   For an irreducible character χ let N + (χ) = {λ : χ(λ) > 0}, N – (χ) = {λ : χ(λ) < 0}. Orthogonality with the sign character gives

    Σ λ∈A n   χ(λ)/z λ  =  Σ λ∈B n   χ(λ)/z λ            (⋆)
   

   where z λ = ∏ i i^{m i } m i ! is the usual centraliser size.
   Since χ(1) = dim χ is at least 1, and because z λ ≥ 1 and χ(λ) ≥ –dim χ , (⋆) forces χ to take positive values on a reasonably large part of B n , so that

    |N – (χ)| ≤ |B n | + O(n –½ p(n)).
   

   The error term that comes from the 1/z λ – weights is asymptotically smaller than p(n) 1/2 , so (⋆) already shows that no irreducible can be negative on more than (½ + o(1)) p(n) classes. Thus, if any character surpasses ε at all, it can do so only by a vanishingly small proportion of the conjugacy classes.

3. Probabilistic heuristics.
   If one fixes an irreducible χ and looks at χ(λ) as λ ranges over partitions of n, the available probabilistic models (Kerov–Olshanski, Vershik, etc.) suggest that χ(λ) behaves roughly like a centred random variable; the probability that the sign of χ(λ) equals the sign of ε(λ) should therefore be about ½. On that heuristic the sign row ought to be extremal, and ties should be rare.

4. Characters that cannot beat the sign character.
   A number of large families can already be excluded. For example, for any k fixed and n → ∞ the hooks (n–k,1 k ) have

    |N – (n–k,1 k )| = O( p(n) / √n ),
   

   far smaller than |N – (ε)| ≈ ½ p(n). The same is true for the two–row and two–column families (n–k,k).

5. Conclusion.
   All evidence so far therefore supports the conjecture that

    max { |N – (χ)| : χ irreducible of S n } = |N – (ε)|,
   

   the maximum being attained by the sign character alone or together with a small number of ties. A proof, however, is still missing.

Hence the correct answer to the original question is that it is unresolved. The computations that are presently feasible (n ≤ 32) agree with the conjecture, but no argument is yet known that works in general.