While not a complete answer, here are some ideas that may be of interest.

In particular, some experimentation shows that $K_L^TK_L$ has $1/2^{L-1}$ on its diagonal, and $1/2^{L}$ and $1/2^{L-1}$ spread out through the matrix in a fairly regular way (one can count nicely). It seems that, for example, in each column of this product, the entry $1/2^{L-2}$ occurs a maximum of $2^{L-1}-1$ times and a minimum of $2^{L-2}$ times. Now we can use the Perron-Frobenius bound on the largest eigenvalue, say $\rho$ of this matrix. For this, first compute the largest possible column sum, which yields
$\rho \le 2^{L-2}\times \frac{1}{2^{L}} +(2^{L-1}-2^{L-2})\times \frac{1}{2^{L-1}} = 3/4$.

While the smallest possible column sum is
$(2^{L-1}-1)\frac{1}{2^{L-2}} + \frac{1}{2^L} = \frac{1}{2}-\frac{3}{2^L} \ge \rho$.

Thus, $\sqrt{\frac{1}{2}-\frac{3}{2^L}} \le \sigma_1(K_L(0)) \le \sqrt{\frac{3}{4}}$,
so that as $L\to \infty$, the lower bound becomes the boring $1/2$. Numerically, it seems that the $\sigma_1(K_L(0))$ goes rather towards $\sqrt{3/4}$, which can probably be characterized more precisely.