First, let me say that since the family $\{g_n\}$ is not only equicontinuous but equilipschitz (that is, there is a Lipschitz constant that works for all these functions at once), we could prove the uniform boundedness of the family in one short statement. I encourage you to look for that proof.
Let us prove something more general:
Proposition: Let $K$ a compact connected space, and $\mathscr{F}$ an equicontinuous family of real-valued (or $\mathbb{R}^n$-valued, we could even state it in more generality) functions. If there is one point $x_0 \in K$ such that the set $\mathscr{F}(x_0) := \{ f(x_0) : f \in \mathscr{F}\}$ is bounded, then $\mathscr{F}$ is uniformly bounded, i.e. there is a $C < \infty$ such that $\lvert f(x)\rvert \leqslant C$ for all $x\in K$ and all $f\in\mathscr{F}$.
Proof: Let $B = \{ x\in K : \mathscr{F}(x) \text{ is bounded}\}$. Then $B$ is open:
Let $x\in B$ and $C(x) = \sup \{ \lvert f(x)\rvert : f\in\mathscr{F}\}$. By the equicontinuity of $\mathscr{F}$, there is a neighbourhood $U$ of $x$ such that $\lvert f(y) - f(x)\rvert \leqslant 1$ for all $f\in\mathscr{F}$ and all $y\in U$. Hence $\lvert f(y)\rvert \leqslant C(x)+1$ for all $f\in\mathscr{F}$ and all $y\in U$, which implies that $U \subset B$, hence $x\in \overset{\Large\circ}{B}$.
Also, $K\setminus B$ is open, for if $\mathscr{F}(x)$ is unbounded, and $U$ a neighbourhood of $x$ such that $\lvert f(y) - f(x)\rvert \leqslant 1$ for all $f\in\mathscr{F}$ and all $y\in U$, then $U\cap B = \varnothing$. For if we had $\lvert f(y) \rvert \leqslant C$ for some $C\in\mathbb{R}$, $y\in U$ and all $f\in \mathscr{F}$, then we'd have $\lvert f(x)\rvert \leqslant C+1$ for all $f\in\mathscr{F}$, and that would contradict $x\in K\setminus B$.
Since evidently $B\cap (K\setminus B) = \varnothing$, and $K$ is connected, we have $B = \varnothing$ or $K\setminus B = \varnothing$. By assumption, $x_0 \in B$, so it follows that $K = B$.
Now $V_n = \left\{ x \in K : \bigl(\exists U \in \mathscr{V}(x)\bigr)\bigl(\forall y\in U\bigr)\bigl(\forall f\in\mathscr{F}\bigr)\bigl(\lvert f(y)\rvert \leqslant n\bigr) \right\}$ for $n \in \mathbb{N}$ is a nested family of open sets that covers $K$, hence there is an $n\in\mathbb{N}$ with $K = V_n$, i.e. $\lvert f(x)\rvert \leqslant n$ for all $x\in K$ and all $f\in\mathscr{F}$.
Between the general theorem and the very short proof from the equilipschitzness of the family, we have the option to use the uniform equicontinuity of the family to give a shorter proof in the given situation, which we sketch:
By the uniform equicontinuity, there is a $\delta > 0$ such that whenever $\lvert x-y\rvert \leqslant \delta$, then $\lvert g_n(x)-g_n(y)\rvert \leqslant 1$ for all $n$. By assumption, there is a $C_0$ such that $\lvert g_n(0)\rvert \leqslant C_0$ for all $n$. Then $\lvert g_n(x)\rvert \leqslant C_0 + 1$ for all $x$ with $\lvert x\rvert \leqslant \delta$. And hence $\lvert g_n(x)\rvert \leqslant C_0 + 2$ for all $x$ with $\lvert x\rvert \leqslant 2\delta$. Generally, $\lvert x\rvert \leqslant k\delta \implies \lvert g_n(x)\rvert \leqslant C_0 + k$. Choose $k \geqslant \frac{1}{\delta}$.
(Expanding on a comment by John Ma.)
The family is uniformly bounded: $|f_n|\le M_0$. It is also equicontinuous, thanks to $|f'_n|\le M_1$:
$$
|f_n(x)-f_n(y)|\le M_1|x-y|
$$
Therefore, there is a subsequence $\{f_{n_k}\}$ converging to a continuous function $f$.
Next step, consider $\{f_{n_k}'\}$ which is also a uniformly bounded equicontinuous family. Extract a further subsequence $f'_{n_{k_l}}$ which converges uniformly. Since $f_{n_{k_l}}\to f$ and $f'_{n_{k_l}}$ converge uniformly, it follows that $f$ is differentiable and $f'_{n_{k_l}}\to f'$.
This process of choosing subsequences continues indefinitely, demonstrating that $f$ is infinitely differentiable. It remains to apply the diagonal argument: let $F_n$ be the $n$th term of the $n$th subsequence. Then $F_n$ converge, with all its derivatives, to the corresponding derivatives of $f$.
Best Answer
A condition on the derivatives cannot guarantee the whole series (take $f_n(x):=(-1)^n$).
However, in the case where $(f_n(x),n\geqslant 1)$ is non-increasing for all $x$, and the sequence is uniformly bounded, this is true (and called Dini's theorem).