The answer is: $C_n \cdot C_n$, where $C_n = \binom{2n}{n}$ is the $n$-th Catalan number.

We divide the proof into several steps. First, some terminology: the last (rightmost) $n$ cells will be called the back cells and the front (leftmost) $n$ cells will be called the front cells. The central, $(n+1)$-st, cell will be called the middle cell.

Claim 1. All the back cells must be removed before any front cell is removed.
Proof. Assume for contradiction that this is not the case. Then there must be a point in time where a front cell is deleted and then immediately after a back cell is deleted. Let us say that the deleted front cell was at position $i$. So all back cells have positions greater or equal to $i+2$. After the cell is deleted all back cells have positions greater or equal to $i+1$. But since we deleted cell $i$, then the total sum is $i$ and this does not increase. So at the next step we delete a cell at position at most $i$, a contradiction. $\square$

Claim 2. The middle cell must contain the number $0$, i.e., $a_{n+1} = 0$.
Proof. Consider the last step in the process where we have total of $2$ cells. One of these is the middle cell, and by Claim 1 the other must be one of the front cells. I.e. we have $(x, a_{n+1})$. On the next move, we remove $x$, which means that $x + a_{n+1} = 1$. So $a_{n+1} = 0$ or $a_{n+1} = 1$. But after that we cannot remove $a_{n+1}$, which means that $a_{n+1} \neq 1$. So $a_{n+1} = 0$. $\square$

Now, let's define a self-destructing sequence to be one with no surviving cells at the end of the process. For example, $(0, 1, 2)$ is self-destructing because $(0, 1, 2) \to (0, 1) \to (1) \to ()$. Let $S_n$ be the set of self-destructing sequences of length $n$. For example, $S_2 = \{(0, 1), (1, 1)\}$. It is clear that the front cells form a self-destructing sequence, i.e., $(a_1, a_2, \dots, a_n) \in S_n$. The back cells also have certain self-destructing quality, which is made more precise in Claim 3 below.

Claim 3. Fix the front sequence $\varphi = (a_1, a_2, \dots, a_n)$. Let $B_\varphi$ be the set of all possible back sequences of length $n$ that can be appended to $\varphi$ (with a $0$ between them) to get a balanced sequence. Then there is a bijection $f: S_n \to B_\varphi$.
Proof. Let $c = n + 1 - \sum_{i=1}^n a_i$ and consider a particular $\sigma = (s_1, s_2, \dots, s_n) \in S_n$. Let $\ell$ be the initial index of the last surviving cell in $\sigma$. Then $f(\sigma) = (s_1, s_2, \dots, s_\ell + c, s_{\ell+1}, \dots, s_n)$ defines a bijection $S_n \to B_\varphi$.

Indeed we claim that the $k$-th deleted cell in $\sigma$ is the $k$-th deleted cell in $\overline{\varphi\ 0f(\sigma)}$ for each $k=1, \dots, n$. Indeed after some deletions let $S$ be the total sum remaining in $\sigma$. Then the total sum remaining in $\overline{\varphi\ 0f(\sigma)}$ is $-\sum_{i=1}^n a_i + 0 + S + c = S + n + 1$. So we delete next the cell in position $S$ in $\sigma$ if and only if we delete the cell in position $S+n+1$ in $\overline{\varphi\ 0f(\sigma)}$.

So $\overline{\varphi\ 0f(\sigma)}$ is clearly a balanced sequence: we first eliminate all cells in the back, then the front. In the same manner it follows that every balanced sequence is of this form. $\square$

So far we have shown that the total number of balanced sequences is $|S_n|^2$. It remains to calculate the size $|S_n|$.

Claim 4. Let $\mathcal{T}_n$ be the set of $2n$-sequences consisting of $n$ zeros and $n$ ones such that in each initial segment the number of $1$'s does not surpass the number of $0$'s. Then $|S_n| = |\mathcal{T}_n|$.
Proof. Let $[n] = \{1, 2, \dots, n\}$, and let us also consider the set $\mathcal{F}_n$ of non-decreasing mappings $f: [n] \to [n]$ such that $f(i) \le i$ for each $i \in [n]$. The claim will follow once we show that $|S_n| = |\mathcal{F}_n|$ and that $|\mathcal{F}_n| = |\mathcal{T}_n|$.

In order to demonstrate that $|S_n| = |\mathcal{F}_n|$, observe that there is an obvious bijective correspondence $a \mapsto f$ between the sets $S_n$ and $\mathcal{F}_n$. Indeed, reversing the self-destructing process for an $n$-sequence $a = (a_1, a_2, \dots, a_n) \in S_n$, simply define $f(i)$ to be the (partial) sum of the existing terms after the $i$-th backward step.

As for $|\mathcal{T}_n| = |\mathcal{F}_n|$, note the following bijective correspondence $t \mapsto f$ between the sets $\mathcal{T}_n$ and $\mathcal{F}_n$. Let $f(i)$ equal $1 + \#(i)$, where $\#(i)$ is defined to be the total number of $1$'s appearing in $t$ before the $i$-th zero.

Finally, it is a known fact that $|\mathcal{B}_n|$ is the $n$-th Catalan number $C_n = \frac{1}{n+1} \binom{2n}{n}$. (The essential idea of the textbook proof of this fact uses the so-called reflection principle of A. D. André.)