This is false in general, even when $A$ and $B$ are fields.

**Example.** Let $A = K = k(t)$, and $B = k(t^{\frac{1}{p^\infty}}) \cong A[\{x_i\}_{i \geq 0}]/(x_0 - t, \{x_{i+1}^p - x_i\}_{i \geq 0})$. Thus,
\begin{align*}
B \otimes_A \bar K &= \bar K[\{x_i\}_{i \geq 0}]/(x_0 - t, \{x_{i+1}^p - x_i\}_{i \geq 0})\\
&\cong \bar K[\{z_i\}_{i \geq 0}]/(z_0, \{z_{i+1}^p-z_i\}_{i \geq 0}),
\end{align*}
through the identification $z_i = x_i - t^{\frac{1}{p^i}}$. This has infinitely many nilpotents $z_i$ (with $z_i^{p^i} = 0$), and clearly there is no single $L$ over which the $z_i$ are defined.

**Remark.** On the other hand, the answer is positive if $f$ is essentially of finite type (i.e. $B$ is a localisation of a finite type $A$-algebra). Indeed, in this case $B \otimes_A \bar K$ is essentially of finite type over $\bar K$, hence Noetherian. Thus, the nilradical $\mathfrak{rad}(B \otimes_A \bar K)$ is finitely generated; say by $x_1,\ldots,x_r$.

Let $L$ be a field over which all the generators are defined, i.e. there exist $y_1,\ldots,y_r \in B \otimes_A L$ such that the image of $y_i$ in $B \otimes_A \bar K$ is $x_i$. Note that each $y_i$ is nilpotent, for example since $B \otimes_A L \to B \otimes_A \bar K$ is injective and each $x_i$ is nilpotent.

Now consider $C = (B \otimes_A L)/(y_1,\ldots,y_r)$. I claim that $C$ is geometrically reduced (as $L$-algebra) (see Tag 05DS for this notion). Indeed, when we apply $- \otimes_L \bar K$, we get $(B \otimes_A \bar K)/(x_1,\ldots,x_r)$, which is reduced by our choice of the $x_i$. $\square$

(As a corollary, we get that $C$ is reduced, so $y_1,\ldots,y_r$ are the only nilpotents in $B \otimes_A L$.)

**Remark.** Note that I never use that $A$ and $B$ are local, Noetherian, regular, etc. It is rather a relative property of the morphism $f$ (in fact, I only need the morphism $f_K \colon K \to B \otimes_A K$).