Limit problem no longer works in Mathematica 11.1.0

I'm struggling to see what the question is, other than "Anyone else experience this problem?", which I guess comes down to, "Can others confirm this is a backslide and that I haven't messed up something?" There is a certain amount of noise surrounding the question: the initial mis-coding of the absolute value and the abuse of PowerExpand in particular. This makes it unclear to me whether more is being sought.

Workarounds

The OP has already noted that coding the source problem (n-th root test) with the usual absolute value Abs works:

Limit[Power[Abs[((1 - 4 n)/(3 n + 2))^(3 n)], (n)^-1], n -> ∞]
(*  64/27  *)

Here is another workaround.

Limit[
 Series[Power[Sqrt[((1 - 4 n)/(3 n + 2))^(3 n)]^2, (n)^-1], {n, ∞, 1}],
 n -> Infinity]
(*  64/27  *)

And another, which perhaps is the best representation of the OP's intended computation bypassing Abs:

Limit[
 Power[Sqrt[(((1 - 4 n)/(3 n + 2))^(3 n))^2], (n)^-1], 
 n -> ∞,
 Assumptions -> n > 0 && n ∈ Integers]
(*  64/27  *)

Problem with PowerExpand

PowerExpand willy-nilly combines the exponents, effectively assuming the bases are positive. That hypothesis is not true in this case, so the result is bogus (due to user error, not an error in Mathematica). Both the incorrect Sqrt[x]^2 and the correct Sqrt[x^2] replacements for Abs[x] for real x expand to the same expression, which has a limit of -64/27 at infinity:

Power[Sqrt[((1 - 4 n)/(3 n + 2))^(3 n)]^2, (n)^-1] // PowerExpand     (* Sqrt[x]^2 *)
Power[Sqrt[(((1 - 4 n)/(3 n + 2))^(3 n))^2], (n)^-1] // PowerExpand   (* Sqrt[x^2] *)
(*  (1 - 4 n)^3/(2 + 3 n)^3  *)

When PowerExpand is applied to Limit, first the limit returns unevaluated, its arguments are expanded, and then the limit it reevaluated and returns the (correct) limit of -64/27 for the expanded function.

Note that because the power Power[z, (n)^-1] is the principal value, the correct and incorrect substitutions for Abs[x] converge to the same value as n -> Infinity.

The fakeabs form Sqrt[x]^2 zigzags because x = ((1 - 4 n)/(3 n + 2))^(3 n) alternates sign and therefore Sqrt[x] alternates real/imaginary.

fakeabs = ReIm@N@Table[Power[Sqrt[((1 - 4 n)/(3 n + 2))^(3 n)]^2, (n)^-1], {n, 150}];
realabs = ReIm@N@Table[Power[Sqrt[(((1 - 4 n)/(3 n + 2))^(3 n))^2], (n)^-1], {n, 150}];
limit = {ReIm[64/27]};    
ListLinePlot[{fakeabs, realabs, limit}, PlotMarkers -> Automatic, 
 Frame -> True, PlotRangePadding -> Scaled[.05], PlotRange -> All, 
 PlotLegends -> {HoldForm@fakeabs, HoldForm@realabs, HoldForm@limit}]

I suppose it is clear that this appears to be a backslide.

Do ComplexExpand to get the right result (Using Version 8.0 )

   Limit[Power[Sqrt[((1 - 4 n)/(3 n + 2))^(3 n)]^2, (n)^-1] // 
            ComplexExpand, n -> Infinity]

    (*  64/27   *)

This has been fixed in 11.2:

Limit[Sqrt[(((1 - 4 n)/(3 n + 2))^(3 n))^2]^(1/n), n -> ∞]
   64/27

Limit[(Sqrt[((1 - 4 n)/(3 n + 2))^(3 n)]^2)^(1/n), n -> ∞]
   64/27

Limit[Abs[((1 - 4 n)/(3 n + 2))^(3 n)]^(1/n), n -> ∞]
   64/27