1. Use the critical points information

Since $f(x)$ is a polynomial of degree $4$, its derivative $f'(x)$ is a polynomial of degree $3$.

Given that the critical points are at $x=-1,0,1$, we must have

$f'(x)=k(x+1)x(x-1)=k(x^3-x)$

for some nonzero constant $k$.

2. Integrate to get $f(x)$

Integrating,

$f(x)=k\int (x^3-x)\,dx = k\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+C$

So,

$f(x)=k\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+C$

where $C$ is a constant.

3. Find $f(0)$

Substituting $x=0$,

$f(0)=C$

We need the set

$T=\{x\in\mathbb R\mid f(x)=f(0)\}$

So solve

$k\left(\frac{x^4}{4}-\frac{x^2}{2}\right)+C=C$

Since $k\neq 0$,

$\frac{x^4}{4}-\frac{x^2}{2}=0$

Multiply by $4$:

$x^4-2x^2=0$

Factor:

$x^2(x^2-2)=0$

Hence,

$x=0,\ \pm \sqrt{2}$

So,

$T=\{0,\sqrt{2},-\sqrt{2}\}$

4. Compute the sum of squares of elements of $T$

The required sum is

$0^2+(\sqrt{2})^2+(-\sqrt{2})^2=0+2+2=4$

5. Compare with the given options

Thus the correct option is

$\boxed{4}$

which is Option D.