[Ljava.lang.String;@ Output Error

I have a piece of code that has what I believe to be an arraylist passed to it in the form of args[].

Nope, you're not dealing with an ArrayList at all but an array. That's what String[] args is declared to be and what it in fact is.

I need to use the sendMessage method to display chat in-game and where I have attempted to create a regular string from my arraylist, what is sent in chat is a '[Ljava.lang.String;@' message instead.

and what you're seeing is the expected result when you call toString() on a String array. Don't do this. Either use java.util.Arrays.toString(myArray) or loop through the array extracting each String item you encounter from the array from inside of the loop.

e.g.,

String msg = java.util.Arrays.toString(args);
OnlinePlayers.sendMessage(ChatColor.GREEN + Player.getName() + ": " + msg);

As an aside, please learn and use Java naming conventions. Variable names should begin with a lower case letter and classes, interfaces, and enums with an upper case letter. This will help prevent confusion about your code.

String Msg = new String (args.toString());

Aside number 2, don't use new String(...) unless you have a strong reason to do so. This will prevent Java from using an already existing String from the String pool, cluttering up your heap with unnecessary objects. A simple foo.toString() is all that would be needed in place of new String(foo.toString()).

Edit regarding:

String NewMsg = null;

String Msg = java.util.Arrays.toString(args);

for( int i = 0; i <= Msg.length() - 1; i++) {

    NewMsg = (args[i] + " ");

}

No, you would want to use java.util.Arrays.toString(...) or a for loop, not both. If you use a for loop your array needs to stay an array. Also, what good is the length of the Msg String as the for loop's max index count?

Instead your logic should be (pseudocode)

Declare a String variable, msg. Assign "" to it.
For int variable i goes from 0 to less than the length of the args array
  concatenate the i'th args item to the msg String.
use your msg String as needed.