A universal operator necessarily generates $\neg x$ for input $x,…,x$
There are two ways to define a universal operator: when constants are allowed, and when they are not allowed. If constants are allowed, then one can define a universal operator which doesn't satisfy the claim, as follows:
$$
T(x,y,z) = \overline{x \land y} \land z.
$$
This is universal since $T(x,y,1)$ is the NAND operator, but $T(0,0,0) = 0$.

Now suppose that constants are not allowed. Since $T$ is universal, there is a $T$-circuit which computes the NOT function $x \mapsto \lnot x$. Let us now consider the possible values that $T(x,\ldots,x)$ takes:


$T(x,\ldots,x) = \lnot x$: in this case the claim holds.
$T(x,\ldots,x) = x$: in this case, induction shows that every $T$-circuit in which the only inputs are copies of $x$ necessarily computes the value of $x$; in particular, no $T$-circuit computes NOT.
$T(x,\ldots,x) = 0$: in this case, when $x = 0$, induction shows that every $T$-circuit always computes 0; in particular, it doesn't compute NOT.
$T(x,\ldots,x) = 1$: similar to the preceding case.


This shows that $T(x,\ldots,x)$ must compute $\lnot x$.