Christoffel symbols vanish in a system of normal coordinates.

Let $p\in \mathcal{M}$, the differential manifold. Let $\mathcal{N}_p$ the normal neighbourhood of $p$. Let $q\neq p$, $q\in \mathcal{N}_p$, be such that $\gamma_v(t)=q$ for some $t$. In normal coordinates (using that there, $\gamma_v (t) = t(v^1 ,...,v^n )$): $$\Gamma^i_{jk}v^jv^k|_q=0$$ But, here, $v$ is fixed, so it can't be concluded that $\Gamma^i_{jk}=0$. But for $p$, it can be applied to every $\gamma_v(t)$ (every geodesic is such that $\gamma_{v'}(0)=p$), i.e., $\forall v$. Then take $v_i=\delta_{ij}$, using the equation above, $$\Gamma^i_{jj}|_p=0 $$ Now set $v_i=(\delta_{ij}+\delta_{ik})$, it follows (using $\Gamma^i_{jj}|_p=0 $) $$\Gamma^i_{(jk)}|_p=0 $$ The symmetric part (assuming $\nabla$ is a general affine connection). So we conclude the symmetric part at $p$ is zero, only at $p$.

One possible route (pretty physicsy, not very diff. geom.-ey, which means it appeals to my way of thinking but probably not yours):

Hint: Consider the local form of geodesics in normal coordinates, and the form of the geodesic equation given this parametrization.

You are correct in saying that knowing $E=G=1, F=0$ is not enough. If we know $E=G=F=0$, then it would be ok because Christoffel symbols then must be 0, but otherwise, it is not enough. Benya and Alvaro's answer was pretty much correct/formally stated, but I think they left out some stepping stones, so I'll try my best to fill them in. So given $\mathcal{N}_{p}$ and your geodesic, $\gamma_{v}(t) = \operatorname{Exp}_{p}(t v)$, because it's a geodesic, it satisfies the geodesic equations: $$\gamma''_v(t)+\Gamma_{i j}^{k} \frac{d \gamma_v^{i}}{d t} \frac{d \gamma_v^{j}}{d t}=0$$ We know that acceleration is 0 since $v \in T_{p} \mathcal{M}$. Therefore, we are left with the following for $v=v^{k} e_{k}$ $$0+\Gamma_{i j}^{k} v^{i} v^{j}=0$$ The rest follows from Benya and Alvaro's answer. Basically if we restrict this to $p$, then $\left.\Gamma_{i j}^{k}\right|_{p}+\left.\Gamma_{j i}^{k}\right|_{p}=0, \text { for all } i, j, k$.

I disagree that knowing $E=G=1$ and $F=0$ is not enough.

I don't recall that Do Carmo provides in explicit formula for the Christoffel symbols, but in a discussion before the Theorema Egregium proof, he implies that such formulae exist. I provide you one here,

$$\begin{bmatrix}E&F\\F&G\end{bmatrix}\begin{bmatrix}\Gamma^u_{uu}\\\Gamma^v_{uu}\end{bmatrix}=\begin{bmatrix}E_u/2\\F_u-E_v/2\end{bmatrix}. $$

By this and the other three formulae, you can conclude every Christoffel symbol vanishes without appeal to geodesics.