You're almost there.

When you execute a query against the Firebase Database, there will potentially be multiple results. So the snapshot contains a list of those results. Even if there is only a single result, the snapshot will contain a list of one result. And your code needs to handle the fact that `` is a list with something like:

roomRef.orderByChild('nickname').equalTo(username).once('value',(snapshot)=> {
  snapshot.forEach((userSnapshot) => {  
    userSnapshot.ref.remove();  
  })
})

The above runs a separate remove() call for each matching node. If you have only one (or a few) nodes with the same name, this is probably fine. But if you have more of them, consider using a single multi-path update to remove all nodes like this:

roomRef.orderByChild('nickname').equalTo(username).once('value',(snapshot)=> {
  let updates = {};
  snapshot.forEach((userSnapshot) => {  
    updates[userSnapshot.key] = null; // 👈 setting to null will remove the node
  })
  roomRef.update(updates); // 👈 send all updates in one call
})

Here snapshot.ref will be roomRef itself therefore using remove() method on that will remove whole allRooms/roomId node. To delete that specific child node, you need a reference to that. Try this:

const roomRef = firebase.database().ref(`allRooms/${rooms}`);

roomRef.orderByChild('nickname').equalTo(`${username}`).once('value').then((snapshot) => {  
  if (snapshot.val()) {
    const keys = Object.keys(snapshot.val())
    snapshot.ref.child(keys[0]).remove().then(() => {
      //         ^^^^^ deleting that child instead of roomRef
      console.log("Data Deleted")
    }).catch(e => console.log(e));  
  }
})