We can define a metric $d$ on the set $\mathcal O(\mathbb C)$ of holomorphic functions: $d(f,g):=\sum_{n=1}^{+\infty}\frac 1{2^n}\min\left(1,\sup_{z\in B_n}|f(z)-g(z)|\right),$ where $B_n=\{z\in\mathbb C, |z|\leq n\}$. The topology given by this metric is equivalent to the topology of the uniform convergence on compact sets. Indeed, if $\{f_k\}$ is a sequence of holomorphic functions on $\mathbb C$ which converges to $f$ on each compact sets, then fix $\varepsilon>0$ and pick $n_0$ such that $\sum_{n\geq n_0+1}2^{-n}\leq\frac{\varepsilon}2$. Since $\{f_k\}$ converges uniformly to $f$ on $B_{n_0}$, we can pick for each $n\leq n_0$ an integer $j_n$ such that for all $k\geq j_n$, $2^{-n}\min\left(1,\sup_{z\in B_n}|f_k(z)-f(z)|\right)\leq \frac{\varepsilon}{2n_0}$. Hence for all $k\geq k_0:=\max_{1\leq n\leq n_0}j_n$, we have $d(f_k,f)\leq\varepsilon$. Conversely, if $d(f_k,f)$ converges to $0$, and $K\subset\mathbb C$ is compact, then take $n$ such that $K\subset B_n$. We have $d(f_k,f)\geq 2^{-n}\min\left(1,\sup_{z\in B_n}|f_k(z)-f(z)|\right)=2^{-n}\sup_{z\in B_n}|f_k(z)-f(z)|$ for $k$ large enough, hence $f_k\to f$ uniformly on $K$.
Now, we prove the sequential continuity of $J$. Let $\{f_k\}\subset\mathcal O(\mathbb C)$ and $f\in \mathcal O(\mathbb C)$ such that $d(f_k,f)\to 0$. Thanks to the Cauchy integral formula we have, if $B(z_0,1)\subset B_N$ $| J(f_k)-J(f)|=|(f_k-f)^{(p)}(z_0)|=\left|\frac {p!}{2\pi i}\int_{C(z_0,1)}\frac{f_k(z)-f(z)}{(z-z_0)^{p+1}}dz\right|\leq p!\sup_{z\in B_N}|f_k(z)-f(z)|,$ hence $\min(1, | J(f_k)-J(f)|)\leq p! 2^Nd(f_k,f).$ Since $d(f_k,f)\to 0$, we have $| J(f_k)-J(f)|\leq 2^{-1}$ for $k$ large enough (say $k\geq k_0$) therefore $\forall k\geq k_0\quad | J(f_k)-J(f)|\leq p!2^Nd(f_k,f),$ which shows that $\lim_{k\to\infty}| J(f_k)-J(f)|=0$.