At each step, four cards are removed from the deck, so a deck is exhausted in $13$ steps.
Let's call that a 'round'.
The game you propose can be restated as follows.
At each round, we draw $13$ ordered cards from the deck, and add up the values of the cards that came before the first ace in our $13$ drawn cards.
If no ace is drawn, we add up the values of all $13$ drawn cards, shuffle them into the remaining $39$ cards (the ones which were not drawn), and repeat the process.
There are $\binom{48}{13}$ sets of $13$ cards which do not contain an Ace, and so there are $\binom{52}{13}-\binom{48}{13}$ sets of $13$ cards which contain at least one Ace.
Hence, the probability that the game ends in a given round is
$$p=1-\frac{\binom{48}{13}}{\binom{52}{13}}=\frac{14498}{20825}\simeq69.62\%$$
Now, if the game has not ended in a given round, then the expected sum of the cards drawn is $13$ times the expected value of a card drawn (by linearity of the expectation).
Since no card drawn was an ace, the expected value of a card drawn is
$$\frac{2+3+4+5+6+7+8+9+10+11+12+13}{12}=\frac{15}2,$$
and hence the expected sum is $l=13\cdot\frac{15}2=\frac{195}2=97.5$.
Now, we need to calculate the expected sum of the cards drawn before the first Ace in a round where the game ends.
Here, we will break the thing into cases.
$\qquad$Number of Aces in cards drawn: $1$
There are $\binom{48}{12}\cdot\binom{4}{1}$ sets of $13$ cards which contain exactly one Ace.
Therefore, given that the $13$ cards drawn contain at least one ace, the probability that we fall in this case (exactly one Ace drawn) is
$$a_1=\frac{\binom{48}{12}\cdot\binom{4}{1}}{\binom{52}{13}-\binom{48}{13}}=\frac{9139}{14498}\simeq 63.04\%$$
The expected position of the lone ace is $\frac{1+2+3+4+5+6+7+8+9+10+11+12+13}{13}=7$, so on average $6$ non-Ace cards will be drawn before it.
The expected sum for this case is hence $s_1=6\cdot\frac{15}2+1=46$.
$\qquad$Number of Aces in cards drawn: $2$
There are $\binom{48}{11}\cdot\binom{4}{2}$ sets of $13$ cards which contain exactly two Aces.
Like before, the probability that we fall in this case is
$$a_2=\frac{\binom{48}{11}\cdot\binom{4}{2}}{\binom{52}{13}-\binom{48}{13}}=\frac{2223}{7249}\simeq 30.67\%$$
Now, things get trickier.
The position of the pairs of aces is a subset of size $2$ on $S=\{1,2,\dots,13\}$, and we are interested in the minimimum of this subset.
Let $X$ denote this random variable.
There are $\binom{13}2$ $2$-subsets of $S$, and only $12$ of them contain the number $1$, which is a guaranteed minimum.
Therefore
$$\mathbb{P}(X=1)=\frac{12}{\binom{13}2}$$
Similarly, there are $11$ $2$-subsets of $S$ whose minimum is $2$.
More generally, for each $k\in\{1,2,\dots,12\}$, $\binom{13-k}1$ of the $2$-subsets of $S$ have minimum $k$, and we find that
$$\mathbb{P}(X=k)=\frac{\binom{13-k}1}{\binom{13}2}.$$
As a sanity check, notice that they add up to $1$.
The expected sum for this case is hence:
$$s_2=\sum_{k=1}^{12}\left(\frac{15}2\cdot (k-1)+1\right) \cdot \mathbb{P}(X=k)=\frac{57}2$$
$\qquad$Number of Aces in cards drawn: $3$
Now we've got most of the work done.
We have that
$$a_3=\frac{\binom{48}{10}\cdot\binom{4}{3}}{\binom{52}{13}-\binom{48}{13}}=\frac{39}{659}\simeq 5.92\%$$
For this case, let $Y$ be the random variable which denotes the minimum of a uniformly sampled $3$-subset of $S$.
Notice that there are $\binom{13}{3}$ $3$-subsets of $S$.
We will have that for each $k\in\{1,2,\dots,11\}$, $\binom{13-k}2$ of the $3$-subsets of $S$ have minimum $k$.
Hence:
$$\mathbb{P}(Y=k)=\frac{\binom{13-k}2}{\binom{13}3}$$
Finally, the expected sum for this case is
$$s_3=\sum_{k=1}^{11}\left(\frac{15}2\cdot (k-1)+1\right) \cdot \mathbb{P}(Y=k)=\frac{79}4$$
$\qquad$Number of Aces in cards drawn: $4$
For this final case we have
$$a_4=\frac{\binom{48}{9}\cdot\binom{4}{4}}{\binom{52}{13}-\binom{48}{13}}=\frac{5}{1318}\simeq 0.38\%$$
and expected sum
$$s_4=\sum_{k=1}^{10}\left(\frac{15}2\cdot (k-1)+1\right) \cdot \frac{\binom{13-k}3}{\binom{13}4}=\frac{29}2$$
Let's put it all together.
Supposing the game ends on a given round, the expected sum for that round will be $($and you can check that the $a_i$ add up to $1)$
$$w=\sum_{i=1}^4a_is_i=\frac{282424}{7249}\simeq38.96$$
Finally, the expected sum for the game will be given by
$$\sum_{n=1}^\infty\,
\underbrace{\mathbb{P}(\text{Game ends on round $n$})}_{(1-p)^{n-1}\cdot p}
\cdot
\underbrace{\mathbb{E}(\text{Value of sum of cards of a game ending on round $n$})}_{(n-1)\,l+w}\\
=\sum_{n=1}^\infty\,(1-p)^{n-1}\cdot p \cdot \Big((n-1)\,l+w\Big)=(w-l)+\frac{l}p
$$
This last step is standard manipulation of series and term-by-term differentiation, but I can further explain if it's not clear.
Therefore, the final answer is
$$\frac{2363461}{28996} \simeq 81.51$$
Here is a basic strategy that should do better than just flipping one card and staying with it:
Always flip a second card after the first. If the second card has a higher value than the first, stick with the second card, otherwise flip the third card.
Why does this better than staying with just the one first card?
Well, as you say, just flipping one card has an expected value of $n+1$
But for the other strategy:
There are 6 cases to consider, each of which is equally likely:
First card is $n$, second is $n+1$. The strategy says to stick with the second card, so outcome is $n+1$
First card is $n$, second is $n+2$. Stick with card. Outcome is $n+2$
$n+1$ followed by $n$. Strategy says to flip third card. Outcome $n+2$
$n+1$ followed by $n+2$. Stick with card: $n+2$
$n+2$ then $n$. Flip third card: outcome is $n+1$
$n+2$ then $n+1$. Flip third card: $n$
Since each of these events is equally likely, the expected outcome of this stratey is just the average of these outcomes, which is $n+\frac{4}{3}$
OK, so that's indeed a better strategy. Is it optimal? I don't know.
Best Answer
You want to take a dynamic programming approach (which I think should be similar to fedja's answer). Let $A(n,\ell,S)$ be the expected reward when the current sum is $n$, the last number drawn is $\ell$ and the set of remaining numbers is $S$. We're looking for $A(0,10,\{1,2,3,4,5,6,7,8,9\})$ (here $\ell$ can be any number greater than $9$).
For each state $A(n,\ell,S)$, we consider either stopping or drawing:
Here is a Python implementation of this logic:
This gives an expected value of $2749727/181440 \approx 15.15502094356261$.
If anyone can work out the formula, we also have the following answers starting with $2$ cards: $5/2, 25/6, 143/24, 463/60, 6853/720, 57229/5040, 66713/5040, 2749727/181440$. (It looks like there could be a $n!$ in the denominator).