**note:** This is not a complete answer, but I think it helps clarify the problem.

Since $K$ is radial, the problem can be formulated in terms of the profile function $\phi$. If we let $r=\|x\|_2$, then the $i$th component of the gradient is
\begin{equation*}
\frac{\partial K}{\partial x_i}
=
\frac{d\phi}{dr}\frac{\partial r}{\partial x_i}
=
\frac{d\phi}{dr}\frac{x_i}{r}.
\end{equation*}
Hence
\begin{equation*}
\|\nabla K(x) \|_2
=
\left| \frac{d\phi}{dr} (r) \right|
\end{equation*}

Using this, the problem can be restated as follows. We want a 1-d function $\phi$ such that
\begin{equation*}
C_d \int_{0}^{\infty} |\phi^{\prime}(r)| r^{d-1} dr
\leq
C
\end{equation*}
for a fixed constant $C$ when
\begin{equation*}
C_d \int_{0}^{\infty} \phi(r) r^{d-1} dr = 1.
\end{equation*}
We could also eliminate the constant $C_d$ (coming from the change to spherical coordinates), and look for a function $\phi$ satisfying
\begin{equation*}
\frac{\int_{0}^{\infty} |\phi^{\prime}(r)| r^{d-1} dr}
{\int_{0}^{\infty} \phi(r) r^{d-1} dr}
\leq
C.
\end{equation*}

As a first attempt at a solution, we could look for a piecewise polynomial
$\phi$. One possible example would be the Wendland function $\phi(r)=(1-r)_+^4(4r+1)$ with $|\phi^{\prime}(r)|=20r(1-r)_+^3$. If we compute the integrals above, we get
\begin{equation*}
\frac{\int_{0}^{1} |\phi^{\prime}(r)| r^{d-1} dr}
{\int_{0}^{1} \phi(r) r^{d-1} dr}
=
\frac{5d(d+5)}{2d+5}.
\end{equation*}
As $d\rightarrow \infty$, so does the fraction. Hence this $\phi$ does not work, and if we look at the integrals, we can see why.

As $d\rightarrow \infty$, the weight function $r^{d-1}$ is weighting the boundary of the support of $\phi$ more heavily. So what we are looking for is a function where $|\phi^{\prime}|$ goes to $0$ at $r=1$ (wlog) as fast as $\phi$. This cannot happen for a polynomial.

Perhaps a Schwartz class $\phi$ is necessary, or maybe a function decaying to zero exponentially fast at $r=1$.