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
$$\frac{2363461}{28996} \simeq 81.51$$
Best Answer
Elaborating on the comment by mlc, let $X_1,\ldots,X_4$ denote the four chosen cards, and let $\text{rank}(X_i)$ denote the rank (i.e., the face value, starting with $1$ for ace and ending with $13$ for king) of the card $X_i$ for $i=1,2,3,4$. The tuple $(X_1,X_2,X_3,X_4)$ is uniformly random from the set of all tuples of four distinct cards. Thus by symmetry, $X_1$ is uniformly random from the set of all $52$ cards (no card is favored more than the others), and therefore $$\mathbb E\ \text{rank}(X_1)=\frac{1+2+\cdots+13}{13}=7.$$ By the same argument, $\text{rank}(X_2),\text{rank}(X_3),$ and $\text{rank}(X_4)$ also have expected rank $7$.
Thus, by linearity of expectation, $$ \mathbb E(\text{rank}(X_1)+\text{rank}(X_2)+\text{rank}(X_3)+\text{rank}(X_4))=7+7+7+7=28. $$