Python 2, 112 bytes

f=lambda n:n>99and f(n/100)+3+f(n%100)-(n%100<1)or n>19and f(n/10)+1+f(n%10)or([0]+[1]*6+[2,1,1,1,3,1]+[2]*7)[n]

Try it online!

Python 2, 84 83 74 67 bytes

lambda n:4*(n>99)+2-n%~9/9-0x55561aaaab/4**(n%100)%4+`n`.count('7')

Thanks to @xnor for golfing off 9 16 bytes!

Try it online!

Python 2, 79 bytes

lambda n:4*(n>99)+([-1]+10*[1]+[3,1]+7*[2]+8*([2]+9*[3]))[n%100]+`n`.count('7')

Straightforward, but longer.

Try it online!

05AB1E, 34 31 bytes

т%U7¢I€Ā`Iт@3*X_(X20@X12Q(X11QO

Try it online or verify all [1,999] test cases.

Explanation:

With all checks mentioned it will result in 1 for truthy and 0 for falsey.

т%         # Take modulo-100 of the (implicit) input

           #  i.e. 710 → 10

  U        # Pop and store it in variable `X`

7¢         # Count the amount of 7s in the (implicit) input

           #  i.e. 710 → 1

I€Ā        # Trutify each digit in the input (0 if 0; 1 otherwise)

   `       # And push all of the mapped values to the stack

           #  i.e. 710 → [1,1,0]

Iт@        # Check if the input is larger than or equal to 100

           #  i.e. 710 → 1 (truthy)

   3*      # Multiply that result by 3 (for 'hund-red and')

           #  i.e. 1 → 3

X_         # Check if variable `X` is 0

           #  i.e. 10 → 0 (falsey)

  (        # And negate that (to remove 'and' when #00)

           #  i.e. 0 → 0

X20@       # Check if variable `X` is larger than or equal to 20 (for '-ty')

           #  i.e. 10 → 0 (falsey)

X12Q       # Check if variable `X` is exactly 12

           #  i.e. 10 → 0 (falsey)

    (      # And negate that (to remove 'teen')

           #  i.e. 0 → 0

X11Q       # Check if variable `X` is exactly 11 (for 'el-ev-en' minus 'one one')

           #  i.e. 10 → 0 (falsey)

O          # Sum everything on the stack (and output implicitly)

           #  i.e. [1,1,1,0,3,0,0,0,0] → 6

JavaScript (ES6), 89 bytes

n=>(s='01111112111312222322',n>99&&+s[n/100|0]+3-!(n%=100))+~~(s[n]||+s[n/10|0]-~s[n%10])

Try it online!

Jelly, 28 25 23 bytes

9ḊŻ;2+⁵Żċ%ȷ2$ạDṠḄƊ+Dċ7Ɗ

Try it online!

How it works

9ḊŻ;2+⁵Żċ%ȷ2$ạDṠḄƊ+Dċ7Ɗ  Main link. Argument: n (integer in [1, ..., 999])



9                        Set the return value to 9.

 Ḋ                       Dequeue; yield [2, 3, 4, 5, 6, 7, 8, 9].

  Ż                      Zero; yield [0, 2, 3, 4, 5, 6, 7, 8, 9].

   ;2                    Concat 2, yield [0, 2, 3, 4, 5, 6, 7, 8, 9, 2].

     +⁵                  Add 10; yield [10, 12, 13, 14, 15, 16, 17, 18, 19, 12].

       Ż                 Zero; yield [0, 10, 12, 13, 14, 15, 16, 17, 18, 19, 12].

         %ȷ2$            Yield n % 1e2.

        ċ                Count the occurrences of the modulus in the array.

                 Ɗ       Combine the three links to the left into a monadic chain.

              D            Decimal; convert n to its array of digits in base 10.

               Ṡ             Take the sign of each decimal digit (0 or 1).

                Ḅ            Convert the array of signs from base 2 to integer.

             ạ           Compute the abs. difference of the results to both sides.

                      Ɗ  Combine the three links to the left into a monadic chain.

                   D       Decimal; convert n to its array of digits in base 10.

                    ċ7     Count the number of 7's.

JavaScript, 86 84 bytes

A tweaked port of TFeld's Python solution.

f=n=>n>99?f(n/100)+3+f(n%=100)-!n:n>19?f(n/10)-~f(n%10):+"01111112111312222322"[n|0]

Try it online

2 bytes saved thanks to Arnauld

Java 11, 105 bytes

n->(""+"".repeat(8)).charAt(n%100)+(n+"").split("7",-1).length-6+(n>99?4:0)

Contains loads of unprintable characters.

Try it online.

Explanation:

n->               // Method with integer as both parameter and return-type

  (""             //  Push string with ASCII-value digits 46666666666867777777

  +"".repeat(8))  //  Appended with 8 times a string with ASCII-value digits 7888888888

  .charAt(n%100)  //  Take the (input modulo-100)'th character of this string (as integer)

  +(n+"").split("7",-1).length

                  //  Count the amount of 7s in the input + 1

  -6              //  Subtract 6 (1 for the 7s+1 count and 5 to map the ASCII-digits to:

                  //              4 → -1; 6 → 1; 7 → 2; 8 → 3)

  +(n>99?         //  And if the input is larger than 99:

     4            //   Add 4

    :             //  Else:

     0)           //   Leave it the same by adding 0

Charcoal, 39 bytes

≔﹪Ｎ¹⁰⁰ηＩ⁻⁻⁻⁺⁺↨Ｅ謬Ｉι²№θ7⁼η¹¹⁼η¹²¬η⁼¹÷ηχ

Try it online! Link is to verbose version of code. This is an arithmetic solution. Explanation:

≔﹪Ｎ¹⁰⁰η

Take the input number modulo 100.

Ｉ⁻⁻⁻⁺⁺

Calculate adjustments to the number of syllables and output the result as a string.

↨Ｅ謬Ｉι²

Start by changing each non-zero digit to 1 and then decoding as base 2. This gives the correct answer for most inputs.

№θ7

Add 1 for each 7.

⁼η¹¹

Add 1 if the number ends in 11.

⁼η¹²

Subtract 1 if the number ends in 12.

¬η

Subtract 1 if the number ends in 00.

⁼¹÷ηχ

Subtract 1 if the number ends in 1X (where X is any digit including 1 and 2).

Perl 5 -pF, 53 bytes

$_=4*/.../+2*/[^0].$/+!/0$/+y/7//-/1[^1]$/-/12$/-/00/

Try it online!

How

$_=4*/.../     # Hundreds place has minimum of 4 sylables (__ HUN-DRED AND),

               # match fails on number <100, and would add 0 here

  +2*/[^0].$/  # Tens place has two syllables if not 0 (__-TY or __TEEN),

               # match fails on numbers <10, and would add 0

  +!/0$/       # Ones place has one syllable if not 0 (__)

               # -- Now adjust for special cases --

  +y/7//       # add a syllable for every 7 present

  -/1[^1]$/    # remove a syllable for 10-19, except 11

  -/12$/       # remove another syllable for 12

  -/00/        # remove the syllable for AND if it's an even hundred

PHP, 158 bytes

<?php $i=fgets(STDIN);for($j=$p=0;$p<strlen($i);){!str_split($i)[$p++]>0?:++$j;};echo substr_count($i,7)+$j+3*($i>=100)-!($i%=100)+($i>=20)-($i==12)+($i==11);

Try it online!

A port of Kevin Cruijssen's solution (unfortunately it doesn't have the same brevity in PHP :) )

-32 thanks to Shaggy!

Wolfram Language 101 Bytes

s = StringSplit;Length[Join@@(WordData[#,"Hyphenation"]&/@ 

Join@@s/@s[IntegerName@#,"-"])]+Boole[#>100]&

Explanation

(substituting StringSplit for s)

Length[Join@@(WordData[#,"Hyphenation"]&/@

Join@@StringSplit/@StringSplit[IntegerName@#,"-"])]+Boole[#>100]&

IntegerName renders the number in American English (i.e. without "and" included in numbers greater than 100.) E.g. 777-> "seven hundred seventy-seven.

StringSplit[IntegerName@#,"-"] removes any hyphens in the rendering.

StringSplit/@ splits the rendering into words.

Join@@ leaves a simple list of words, without embedded list (in the case that a hyphen appeared).

WordData[#,"Hyphenation"] breaks up a single word into its syllables.

Join@@ leaves a simple list of syllables in all of the words.

Length counts the syllables

+Boole[#>100] adds 1 to the syllable count for those numbers greater than 100 (because of the additional "and" employed in the British English rendering).