a possible approach using reduce and Set. for the accumulator I'm initializing it with the first array (x.c). Doesn't matter which one since in the end we are taking intersection anyway. Inside the reduce I'm intersecting the accumulator with the current iteration array

const x = {
  c: ['Full_Name', 'Last_Name', 'Email', 'Organizations', 'MISC', 'UNIQUEID'],
  small: ['Title', 'Organizations', 'URL', 'UNIQUEID'],
  big: ['abc', 'Organizations', 'def', 'UNIQUEID']
}


const res = Object.values(x).reduce((acc, curr) => {
  return new Set(curr.filter(i => acc.has(i)))
}, new Set(x.c))

console.log([...res])

almost similar with a for loop except I'm skipping first element in loop

const x = {
  c: ['Full_Name', 'Last_Name', 'Email', 'Organizations', 'MISC', 'UNIQUEID'],
  small: ['Title', 'Organizations', 'URL', 'UNIQUEID'],
  big: ['abc', 'Organizations', 'def', 'UNIQUEID']
}

const values = Object.values(x)
let acc = new Set(values[0])
for (let i = 1; i < values.length; i++) {
  acc = new Set(values[i].filter(j => acc.has(j)))
}

console.log([...acc])

You're looking for a Set intersection -

function intersection2(a, b) {
  if (a.size > b.size)
    return intersection2(b, a)
  const r = new Set
  for (const v of a)
    if (b.has(v))
      r.add(v)
  return r
}

You can compute the intersection of many sets by combining reduce and intersection2 -

function intersection(s, ...sets) {
  return sets.reduce(intersection2, s)
}

To use it with your code, get all values of x using Object.values and map over them to create an array of sets. Finally call intersection with all of the sets -

const x = {
  c: ['Full_Name', 'Last_Name', 'Email', 'Organizations', 'MISC', 'UNIQUEID'],
  small: ['Title', 'Organizations', 'URL', 'UNIQUEID'],
  big: ['abc', 'Organizations', 'def', 'UNIQUEID']
}

const result =
  intersection(...Object.values(x).map(v => new Set(v)))

console.log(result)

Set [ "Organizations", "UNIQUEID" ]

If you want the result as an array, you can convert it using Array.from -

console.log(Array.from(result))

[ "Organizations", "UNIQUEID" ]