Not exactly an answer, but too long for a comment.

Assume there is such approximation to the identity. We may define a linear form
$u\colon L^1(T)\to C$ by
$$u(f)=\lim_{\varepsilon\to0} f*k_{\varepsilon}(0).$$

This linear form will be continuous when restricted to $L^\infty(T)$
because for $f\in L^\infty(T)$
$$|u(f)|=\lim_{\varepsilon\to0} |f*k_{\varepsilon}(0)|=
\lim_{\varepsilon\to0} \Bigl|\frac{1}{2\pi}\int_{T}f(t)k_\varepsilon(-t)\,dt \Bigr|\le \Vert f\Vert_\infty.$$
Also for a continuous $f$ we will have $u(f)=f(0)$.

A continuous linear form in $L^\infty(T)$ is associated to a finitely additive
measure $\mu$ on the measurable sets of $T$ by
$$u(f)=\int_{T}f(t)\mu(dt)$$
If $\mu$ were countably additive it will be the Dirac delta (this is the only
countably additive measure that gives integral $f(0)$ when $f$ is continuous),
but this is not true
because if $M$ is of Lebesgue measure $0$ we have $u(\chi_M)=u(0)=0$ and there
is no problem assuming that $0\in M$.

These finitely additive measures defined on the measurable sets of $T$
vanishing on sets of measure zero but not countably additive can be proved to
exists, but they are similar to non-measurable sets (I will call them phantoms).
So they can not be so easily defined.

The hyperplane $u^{-1}(0)$ of $L^1(T)$, is

definable in terms of a sequence of ordinals. I think that in Solovay's model of ZFC,
it can be proved that, in this case, it is closed.
This means $u$ is continuous. But this is clearly
false because then there will exists a function $g\in L^\infty(T)$ such that
$$u(f)=\frac{1}{2\pi}\int_{T}f(t)g(t)\,dt.$$

So it is consistent with ZFC that there is no such approximation of the identity.

But I have forgotten now the details to show that in Solovay's model an hyperplane
definable from a sequence of ordinals is closed. Perhaps it was that a projective
hyperplane was closed. Perhaps somebody here can close my argument.