How to insert data into a database which uses the same input name variable for multiple rows in a table?

Assuming the id field is an auto-incrementing primary key of your Invoices table: Leave that one, when inserting. Your current code always tries to insert 1 or 2 as id for every request.

If its not an auto-increment you need to generate a valid id in your app, $i alone won't do the trick.

foreach ($_POST['item_name'] as $index => $itemName) {
    $quantity = $_POST['quantity'][$index];
    $price = $_POST['price'][$index];

    // DONT DO THIS! Its prone to SQL injection 
    $sqlConnect->query("INSERT INTO `Invoices` (`item_name`, `quantity`, `price`) VALUES ('".$itemName.
        "', '".$quantity.
        "', '".$price.
        "')");
}

Not sure what your db-connection library is, but currently your code is prone to SQL injection attacks. You should use prepared statements and parameterization. Please see what Bobby Tables says.

To have a real dynamic form, where undefined number of items can be entered, you need JavaScript. but for now, I would suggest to change your form to something like this:

<input name='items[1][name]'>
<input name='items[1][quantity]'>
<input name='items[1][price]'>
<!-- and for second item: -->
<input name='items[2][name]'>
<input name='items[2][quantity]'>
<input name='items[2][price]'>

so your $_POST data structure will be like this:

items:[
        [
          quantity:
          price:
          name: 
        ],
        [
          quantity:
          price:
          name: 
        ],
]

This way, you can loop through items, not item properties. and your loop will be like this:

if (isset($_POST['submit'])) {
    var_dump($_POST);
    foreach($_POST['items'] as $item){
         $sql = 'INSERT INTO invoices (name, quantity, price)
                 VALUES ($item['name'], $item['quantity'], $item['price'])';
         $sqlConnect->query($sql);
    }
} else {
    die('No item data to process');
}

Note that in MySQL you can set id field to auto_increment, so no need to include it in your query.

Also, use var_dump() to know what you are dealing with, like var_dump($_POST);