This is one of many other solutions you can do

const a = [
[1,1,2,1,3], 
[2,1,4,4,5],
[3,1,4,5,1]
];

// flat the two dimensions array to one dimension array
const transformedArray = a.flat();

const unifiedArray = [];
const countDuplicates = [];

// create a new array containing unique values
// you can use the Set(..) object for this too
transformedArray.forEach(elem => {
    if(!unifiedArray.includes(elem)) {
    unifiedArray.push(elem);
  }
});

// start counting
unifiedArray.forEach(elem => {
    const countElem = transformedArray.filter(el => el === elem).length;
  countDuplicates.push({
    element: elem,
    count: countElem
  });
});


// test values
console.log(transformedArray);
console.log(countDuplicates);

At first I though to concatenate the arrays into 1 big one, but I assume you want the duplicates that appear in all arrays so:

var arr1 = [1, 2, 3, 4]
var arr2 = [1, 2, 4, 8]
var arr3 = [2, 4, 6, 8]
var duplicates = [];
arr1.forEach(function(item) {
  // you could change the && to || if you want
  if ((arr2.indexOf(item) > -1) && (arr3.indexOf(item) > -1)) {
    duplicates.push(item)
  }
})


console.log("" + duplicates)

this is simple solution of your question

const arr1 = [2, 3, 4, 5, 6, 7, 8];
const arr2 = [1, 2, 3, 9];
const arr3 = [11, 2, 3, 12, 13];

const duplicates = arr1.filter((i) => arr2.includes(i) && arr3.includes(i));

console.log(duplicates);