How to get rid of the error message when evaluating a highly oscillatory numerical integral?

The shortest and best way between two truths of the real domain often passes through the imaginary one.

— Jacques Hadamard

By taking a complex path, I get the answer without any complaints from Mathematica.

parabolic[a_, x_] = Simplify[InterpolatingPolynomial[{{-10, 0}, {0, a}, {10, 0}}, x]]

With[{a = 1}, 
     Re[NIntegrate[With[{x = x + I parabolic[a, x]}, 
                        Exp[-x^2 + 100 I x]] (1 + I Derivative[0, 1][parabolic][a, x]),
                   {x, -10, 10}, WorkingPrecision -> 20]]]
   5.1113608752199029964*10^-46

If you want further reading, see this or this.

You can try setting the AccuracyGoal lower than WorkingPrecision, which yields the correct result without a reported warning.

NIntegrate[Exp[-x^2] Cos[100 x], {x, -10, 10}, 
 Method -> {"LevinRule"}, WorkingPrecision -> 50, AccuracyGoal -> 35]

5.1113608752199120138254477520179596033660767259737*10^-46

NIntegrate[Exp[-x^2] Cos[100 x], {x, -10, 10}, 
 Method -> {"LevinRule"}, WorkingPrecision -> 100, AccuracyGoal -> 90]

5.11136087521990299641307044175882379310640009556142432636206474874790\ 3858004996545055056958734154112*10^-46

Reference.

When you give a setting for WorkingPrecision, this typically defines an upper limit on the precision of the results from a computation. But within this constraint you can tell the Wolfram Language how much precision and accuracy you want it to try to get.

In a highly oscillatory function, the last few digits of precision may not converge and/or take extended time.

When in doubt, use brute force.

-- Ken Thompson

AbsoluteTiming[
 NIntegrate[Exp[-x^2] Cos[100 x], {x, -10, 10}, 
  Method -> {"LocalAdaptive", "SymbolicProcessing" -> 0, "SingularityHandler" -> None}, 
  MaxRecursion -> 100, 
  AccuracyGoal -> 90, WorkingPrecision -> 100]
]

(* {100.201, 
 5.111360875219902996413070441758823793106400095561424326362063281173449212599595171095728154857298543*10^-46} *)