JavaScript (Node.js), 41 bytes

f=x=>/^".*"$/.test(x)?f(eval(x)):x.length

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Trivial

Jelly, 11 bytes

ŒV.ị=”"ȦƲ¿L

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-2 bytes thanks to Jonathan Allan

Explanation

ŒV.ị=”"ȦƲ¿L    Main Link
  ------Ʋ      Link Grouping (see below)
         ¿     While
  .ị               the first and last (index at 0.5)
       Ȧ           are all
    =”"            equal to "
ŒV             Evaluate (Python)
          L    Length

The footer runs all tests and prints 1 for any correct answers. Ỵœṣ“ -> ”$€Ç=V}ɗ/€ means "split on newlines, split each string on substrings equal to " -> ", then for each pair, check that the result of the main link on the left side equals the right side evaluated (to convert to a number)".

For the link grouping, Ʋ combines four links into a monad, so you'd think we'd need to use $ to capture the fifth, but since ”"Ȧ would be an LCC (a chain that begins with a nilad and only has monads and dyad-nilad/nilad-dyad pairs after it, so a chain with a constant result), it doesn't count that as a link and therefore keeps going one extra time. This is an important tip when understanding the link combining quicks as all of them do this and it is easy to miscount or misunderstand the grouping because it may be taking more links than you expect.

ŒV.ị⁼⁾""Ʋ¿L would also work where instead of =”"Ȧ to check that all are equal to ", we just check that it's equal to ['"', '"']. This also makes the grouping a bit simpler because we don't have any potential LCCs so it just takes four links.

Retina 0.8.2, 47 bytes

+`(?=.*"$)(^"(?!$)|\\(.)|(?!^)"$)(?<=^".*)
$2
.

Try it online! Link includes test cases. Explanation:

(?=.*"$)

Ensure the string ends with a " before making any replacements.

(^"(?!$)|\\(.)|(?!^)"$)

Replace a leading ", an escaped character or a trailing ", but not a lone ".

(?<=^".*)

Ensure that the string starts with a " before making any replacements.

$2

Unescape the character.

+`

Repeat until the string can't be unquoted further.

.

Get the length of the final string.

If " is excluded as being an unsupported input, then ten bytes can be saved by removing (?!$) and (?!^).

Excel ms365, 138 bytes

Assuming input in A1:

=LET(x,LAMBDA(f,s,IF(s<"",s,f(f,IFNA(SUBSTITUTE(SUBSTITUTE(MID(s,XMATCH("""*""",s,2)+1,LEN(s)-2),"\\","\"),"\""",""""),LEN(s))))),x(x,A1))

It's a recursive LAMBDA that will keep calling itself untill no more starting+leading double quotes.

Google Spreadsheets, 107 bytes

Again, assuming input in A1, applying the same recursive logic with a few slight changes making use of the regex-functions:

=LET(x,LAMBDA(f,s,IF(REGEXMATCH(s,""".*"""),f(f,REGEXREPLACE(s,"^""|""$|\\(\\|"")","$1")),LEN(s))),x(x,A1))

Perl 5 -pl, 45 bytes

s/\\(\\|")/$1/g while s/^"(.*)"$/$1/;$_=y///c

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Perl 5 -pl, 54 bytes

Handles the "\"foo\\\"" testcase I proposed in the comments. If that's not a valid test case, then the shorter version above will suffice.

s/\\(\\|")/$1/g while s/^"(.*[^\\])"$/$1/;say;$_=y///c

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Python, 51 52 bytes

f=lambda s:f(eval(s))if'"'==s[-1:]==s[:1]else len(s)

Not complicated, but it works well.

Explanation:

Define a function f that takes an argument s:

f=lambda s:

If the string begins and ends with ", return the value of the function with backslashes evaluated:

f(eval(s))if'"'==s[-1:]==s[:1]

Otherwise, output the length of the string:

len(s)

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Charcoal, 34 33 bytes

Ｗ¬∨⌕θ"⌕⮌Φθλ"≔⪫⁻⪪✂θ¹±¹¦¹\\¦\¦\θＩＬθ

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

Ｗ¬∨⌕θ"⌕⮌Φθλ"

Repeat while the string starts with " and ends with a different "...

≔⪫⁻⪪✂θ¹±¹¦¹\\¦\¦\θ

... remove those "s, split the string on \\, remove any remaining \s, then join the string on \, thus unquoting the string.

ＩＬθ

Output the length of the final string.

(Yes I could use eval to save 14 bytes but that's boring.)

If " is excluded as being an unsupported input, then two bytes can be saved by replacing Φθλ with θ.

Python 3, 9190 bytes

f=lambda s:f(s[1:-1].replace('\\"','"').replace(r'\\','\\'))if'"'==s[-1:]==s[0]else len(s)

because eval is evil :P, with inspiration from Jakav

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QBASIC, 302 bytes

Q$ = CHR$(34): LINE INPUT T$: DO WHILE LEFT$(T$, 1) + RIGHT$(T$, 1) = Q$ + Q$: U$ = MID$(T$, 2, LEN(T$) - 2): T$ = "": C$ = "": FOR J = 1 TO LEN(U$): C$ = C$ + MID$(U$, J, 1): B = -(C$ <> "\"): T$ = T$ + RIGHT$(C$, B * ((INSTR("\\" + Q$, C$) > 0) + 2)): C$ = LEFT$(C$, 1 - B): NEXT: LOOP: PRINT LEN(T$)

A bit of nostalgia...

Explanation:

'Set quote mark because QBASIC doesn't have string escaping 
'then get string
Q$ = CHR$(34)
LINE INPUT T$
'Process the string while it starts and ends with quotes
DO WHILE LEFT$(T$, 1) + RIGHT$(T$, 1) = Q$ + Q$
    'Strip enclosing quotes, assign to a temporary variable,
    'prepare to rebuild the original string,
    'and prepare to track current character(s)
    U$ = MID$(T$, 2, LEN(T$) - 2)
    T$ = ""
    C$ = ""
    'Process string character by character.
    FOR J = 1 TO LEN(U$)
        'Add current character to current status
        C$ = C$ + MID$(U$, J, 1)
        'QBASIC boolean ops return -1 for true, 0 for false,
        'so set B to 0 if current character is an escape char, to 1 otherwise.
        B = -(C$ <> "\")
        'Add current character to string if not an escape char,
        'de-escaping if appropriate
        T$ = T$ + RIGHT$(C$, B * ((INSTR("\\" + Q$, C$) > 0) + 2))
       'Clear current character unless it's an escape char
        C$ = LEFT$(C$, 1 - B)
    NEXT
LOOP
PRINT LEN(T$)

05AB1E, 10 bytes

Δ¬'"Qi.E]g

Try it online or verify all test cases. (Note: in the single TIO, the input is wrapped within """-quotes to always have a string input.)

Explanation:

Δ         # Loop until the result no longer changes:
 ¬        #  Push its first character (without popping the string)
  '"Qi   '#  If it's a double quote:
      .E  #   Evaluate the string as Elixir code
]         # Close both the if-statement and changes-loop
 g        # Pop and push the length of the reduced string
          # (which is output implicitly as result)

Pyth, 20 bytes

L?&b&qhbNqebNyvblbyw

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Same approach as Jakav's Python solution.

Vyxal l, 70 bits^v2, 8.75 bytes

Sλh\"=ßE;Ẋ

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Bitstring:

0110001110110011010000001101101101100000011010011000101001000111010000

Port of 05ab1e except i had to add a "stringify" at the start for the final case