JavaScript (ES7), 146 bytes

s=>s.replace(/(AA )?A+(,)?/g,(s,P,S)=>x=["32_10+-&*^"[s.length+4*!P-4*!S],...x],x=[])&x.map(o=>x.push(v=1/o?o:(eval(x.pop()+o+x.pop())**2)**.5))|v

Try it online!

Method

We first split the input string into tokens, using the following regular expression:

/(AA )?A+(,)?/g
 \___/ \/\_/
   |    | '--> S = optional suffix \
   |    '----> main part            |--> s = full match
   '---------> P = optional prefix /

We turn each token into an integer in the range \$[0\dots9]\$ by applying the formula:

s.length + 4 * !P - 4 * !S

We translate this integer back into an output token, which is either a digit or a JS operator.

This gives:

We store these tokens in reverse order in an array x[] and iterate on this array to evaluate the expression:

x.map(o =>          // for each value o in x[]:
  x.push(           //   push in x[] (reused as a stack):
    v =             //     store in v
    1 / o ?         //     if o is a digit:
      o             //       use o
    :               //     else (o is an operator):
      (             //
        eval(       //       evaluate:
          x.pop() + //         the 1st operand
          o +       //         followed by the operator
          x.pop()   //         followed by the 2nd operand
        ) ** 2      //       compute the absolute value
      ) ** .5       //       (this is shorter than Math.abs())
  )                 //   end of push()
)                   // end of map()
| v                 // return the last result

Charcoal, 98 bytes

Ｆ⮌⪪⪫⪪⪫⪪Ｓ×A³,,¦AA ¦,¦ ≡ι,A,⊞υ⁺⊟υ⊟υ,AA,⊞υ↔⁻⊟υ⊟υ,,,⊞υ×⊟υ⊟υ,,,,⊞υ＆⊟υ⊟υ,,A,⊞υΣ⁻…⮌υ²＆⊟υ⊟υ⊞υ﹪⌕⪫,A⪫,,ιA⁴Ｉυ

Try it online! Link is to verbose version of code. Explanation:

Ｆ⮌⪪⪫⪪⪫⪪Ｓ×A³,,¦AA ¦,¦ ≡ι

Take the input string, replace AAA with ,, and AA with ,, then split on spaces and switch over the tokens in reverse order.

,A,⊞υ⁺⊟υ⊟υ

AA A, becomes ,A,; Plus the top two elements of the stack.

,AA,⊞υ↔⁻⊟υ⊟υ

AA AA, becomes ,AA,; Absolute Minus the top two elements of the stack.

,,,⊞υ×⊟υ⊟υ

AAA, becomes ,,,; Times the top two elements of the stack.

,,,,⊞υ＆⊟υ⊟υ

AA AAA, becomes ,,,,; BitwiseAnd the top two elements of the stack.

,,A,⊞υΣ⁻…⮌υ²＆⊟υ⊟υ

AAAA, becomes ,,A,, which has to be emulated as Charcoal does not have the XOR operator.

⊞υ﹪⌕⪫,A⪫,,ιA⁴

Otherwise, work out the value of the constant and push that to the stack.

Ｉυ

Output the final value on the stack.

05AB1E, 54 52 44 bytes

ðì•9X':;üÞVb•…,A ÅвJ•ZΩнG•S£"^&*α+0132"S:R.V

Try it online or verify all test cases.

Explanation:

ðì            # Prepend a space to the (implicit) input-string
•9X':;üÞVb•  '# Push compressed integer 163244455251778732342
 …,A Åв       # Convert it to custom-base ",A ",
              # aka, convert to base-3 and index into ",A "
       J      # Join this character-list together to a string:
              #  "AAAA, AA AAA, AAA, AA AA, AA A,AAAAAAAAA AA"
•ZΩнG•        # Push compressed integer 585764341
      S       # Convert it to a list of digits
       £      # Split the earlier string into parts of these sizes:
              #  ["AAAA,"," AA AAA,"," AAA,"," AA AA,"," AA A,","AAAA","AAA","AA A","A"]
"^&*α+0132"   # Push string "^&*α+0132"
           S  # Convert it to a list of characters
            : # Replace the earlier substrings to these characters
R             # Reverse the entire string
 .V           # Evaluate it as 05AB1E code
              # (after which the result is output implicitly)

See this 05AB1E tip of mine (section How to compress large integers?) to understand why •9X':;üÞVb• is 163244455251778732342 and •ZΩнG• is 585764341.

05AB1E executes its stack in Reverse Polish notation, hence the need for the R and why .V works on the (transliterated) string as a whole without any other modifications.