So I was thinking of doing this recursively: $f(x,i)$ is equal to the probability of rolling greater than $x$ and landing on $i$ on the last roll. $f(0,i) = 1/6$ for $i = \{1,2,..,6\}$. $f(1,i) = 1/6 + 1/6f(0,i)$ for $i = \{2,...,6\}$ and $f(1,1) = 1/6f(0,1)$. Finally, we list out this recursion until we get $f(13,i)$ and see for what value of $i$ is $f$ the largest.
Is there a better way to approach this or an easy way to simplify this method?
We are not asked the exact probabilities $f(13,i)$, but it is somehwat obvious that for $n\gg 0$ we have $f(n,i)\sim i$ (and hence $f(n,i)\approx \frac i{21}$). So even without calculation it is reasonable to assume that $f(13,6)>f(13,i)$ for all $i\ne 6$.
And indeed, any sequence of rolls that ends with $i$ exceeding $13$ can be mapped to a sequence of rolls that ends in a $6$ exceeding $13$ (and having the same rolls before). Therefore $f(13,6)\ge f(13,i)$ for all $i$. Now note that any sequence ending exactly at $13$ with an $i<6$ can be turned to a sequence exceeding $13$ and ending in $6$, we see that in fact $f(13,6)>f(13,i)$. (In this last step we used that $13\ge i$ for all $i<6$ so that the existence of a sequence summing to exactly $13$ with an $i$ as last roll is guaranteed).
The answer to your particular problem (which is the most probable last dice) is already given. Regarding the more general problem (computing $f(n,i)$), we can see that
$$f(n,i)= \frac{1}{6}\sum_{k=0}^{i-1} S(n-k)$$
where $S(m)$ is the probability that an unbounded running sum of dice hits the value $m$. Now, surely that problem must have already been analyzed in a fantastic math site I know... let me see... yes, here it is. We see that $S(m)$ has no simple closed formula, so your recursive approach is reasonable. And we can also intuitively see that $S(m)$ must asympotically tends to a constant (specifically, to 2/7), so $f(n,i) \approx i/21$ for large $n$.
Denote by $p_r(n)$ the probability that the last roll is $r$, when $n$ more points are needed to end the game. Then $$p_r(1)={1\over6}\quad(1\leq r\leq6)\ ,$$ and for convenience we put $p_r(n):=0$ when $n\leq0$. It is then easy to see that the $p_r(n)$ for each $r\in[6]$ satisfy the recursion $$p_r(n)={1\over6}\left(\sum_{k=1}^6 p_r(n-k) + 1_{1\leq n\leq r}\right)\qquad(n\geq1)\ .\tag{1}$$ For $n>r$ only the homogeneous part $$p_r(n)={1\over6}\sum_{k=1}^6 p_r(n-k)$$ remains. Its characteristic polynomial $\chi(\lambda)=6\lambda^7-7\lambda^6+1$ has a double zero at $1$ and five zeros of absolute value $<1$. It follows that there are constants $a$, $b$ with $$p_r(n)=a n + b +o(1)\qquad(n\to \infty)\ .$$ As $0\leq p_r(n)\leq 1$ for all $n$ we necessarily have $a=0$; but $b$ has to be determined from the initial conditions $\bigl(p_r(n)\bigr)_{r-5\leq n\leq r}$, which result after applying $(1)$ $r$ times.
Running $(1)$ on the computer corroborates the following conjecture: $$\lim_{n\to\infty} p_r(n)={r\over21}\ .$$ Heuristically this can be explained as follows: A roll of $r$ lets the sum jump over $r$ half integers. Since all rolls $r\in[6]$ are equiprobable and $\sum_{r=1}^6 r =21$, on average ${r\over21}$ of all half-integers are jumped over by a roll of $r$. When an $N\gg1$ is chosen then the probability that the mark at $N+{1\over2}$ is jumped over by a roll of $r$ is therefore equal to ${r\over21}$.
Professional random walkers should be able to prove this rigorously.
