Not 1, not 2, not 3, not 4, but 5!

We can use modulo $8$ to solve this problem.

In modulo $8$ , if $x$ is an even number, $x^{12} = x^{10} = x^8 = x^6 = x^4 = 0(mod 8)$ .

In the other hand if $x$ is an odd number, $x^{12} = x^{10} = x^8 = x^6 = x^4 = 1(mod 8)$ .

When $n > 2$ , $2^n = 0(mod 8)$ and $2^n - 1 = -1(mod 8) = 7(mod 8)$

Using these facts we can now solve the problem.

In modulo $8$ , $a^{12} + b^{10} + c^8 + d^6 + e^4$ can only be equal to $0$ , $1$ , $2$ , $3$ , $4$ , and $5(mod 8)$ .

Since $2^n - 1 = 7(mod 8)$ , $n$ cannot be greater than $2$ .

If $n$ is equal to $0$ , $1$ , or $2$ , there will be ordered quintuples $(a, b, c, d, e)$ that will satisfy the equation.

Example: If $n = 0$ , $(a, b, c, d, e) = (0, 0, 0, 0, 0)$

If $n = 1$ , $(a, b, c, d, e) = (0, 0, 0, 0, 1)$

If $n = 2$ , $(a, b, c, d, e) = (0, 0, 1, 1, 1)$

Therefore, $n = 0, 1, 2$ are the only possible values of $n$ .

Since there are $2014$ nonnegative integers less than $2014$ and $3$ of them are the only possible values of $n$ . Therefore there are $2014 - 3 = \boxed{2011}$ nonnegative integer values of $n$ such that there will be no possible ordered quintuples of nonnegative integers $(a, b, c, d, e)$ that will satisfy the equation.