The maximum period is
5460
Warm-up
An exchange instruction swaps two positions in line. Any permutation of the 16 positions can be achieved by a dance program with only exchange instructions, by decomposing the permutation into swaps.
To find the period of a dance program with only exchange instruction, note that it breaks down into cycles that split up the 16 positions. Repeating the program some number of times returns us to the original dance line only when each cycle has returned to the start, and so after a number of steps that's a multiple of all these cycle sizes. The period is therefore the least common multiple of the cycle sizes, which form a partition of 16. The largest possible period for 16 is 140 according to this OEIS sequence, achieved as $\textrm{LCM}(4,5,7)$.
Combining instructions
But, we can do better with a dance program with both exchange and partner instructions.
Note that any partner instruction commutes with any exchange instruction -- swapping positions and swapping dancer names acts independently, and you can think of them as left- and right- compositions on the mapping given by the line.
So, we can simplify any dance program to group all exchange instructions together, followed by all partner instructions. Likewise, repeating this program $k$ times is the same as first doing all of its exchange instructions $k$ times, then its partner operations $k$ times.
We will choose the exchange permutation and the partner permutation to have individually large periods that are relatively prime to each other.
No short-circuiting
We show that the overall period of the dance program is then the product of these two periods. One might worry the overall period is smaller, since it's possible to "short-circuit" and return to the initial state without the exchange permutation or the partner permutation being the identity, but acting on the dance line as the identity in combination. But, we show this won't happen here.
Suppose that after $k$ repetitions of the dance program, the dance line is back as it started. Then, this is also the case after $ka$ repetitions, where $a$ is the period of the exchange permutation. Since $ka$ is a multiple of $a$, after $ka$ repetitions the exchange permutation is the identity. But this means the partner permutation must also be the identity since the whole dance line is back to where it started. Naming its period $b$, this means $ka$ is a multiple of $b$. But since $a$ and $b$ are relatively prime, this means the period $k$ is a multiple of $b$. A similar argument shows that it's a multiple of $b$, and therefore of $ab$. So, $k$ must be at least $ab$, and therefore the period is $ab$.
Largest overall period
So, we want to find two permutations of 16 elements, so that the product of their periods is as large as possible, and these periods are relatively prime. Recalling that the period of a permutation is the LCM of its cycle lengths, which are a partition of its 16 elements, we want to split two copies of 16 into relatively prime values whose product is as large as possible.
We can consider just cycle lengths that are powers of a prime, since any cycle length $xy$ with relatively prime $x$ can be split into two cycles $x$ and $y$, using up $x+y$ elements which is less than the $xy$ before. Note that we can always "pad" to a higher value by putting unused elements into their own cycles of length 1.
The usable powers of a prime that are at most 16 are: $2, 4, 8, 16, 3, 9, 5, 7, 11, 13$. We're allowed 32 in total, and want a product as big as possible. We can only pick one power of 2 and one power of 3.
In general, picking small numbers is more efficient than picking large ones -- if we think of additive contributions to the log of the product, the cost-benefit ratio of $n$ as $\frac{\log n}{n}$. We can similarly consider the marginal contribution of, say, changing 4 to 8 as doubling for the cost of 4. We can then sort the prime powers starting from the most f efficiency, treating powers of 2 and 3 as the marginal benefit over the previous power:
$2, 4, 3, 5, 7, 11, 13, 9, 8, 16$
A good heuristic here is taking the best ones from the start until we run out of total value of 32, giving $4,3,5,7,11$ for a total of 30. With 2 left we can do no better than bump up 11 to 13, giving $4,3,5,7,13$. I think one can prove this is optimal since it uses the most efficient possible values except for replacing 11 with 13 which is unavoidable.
We also need to make sure the total of 32 can be split into two groups adding to 16, and indeed it can: $(4,5,7)$ and $(3,13)$. (In fact, $(4,5,7)$ was the optimal single-permutation split from the warm-up.) These have products of 140 and 39, for an overall period of 5460
Creating the dance programs
To finish, let's step back and see how one would create a dance program with this period. We want our exchange instructions to split into cycles of $(4,5,7)$. We could do this, say, but having the first 4 positions rotate in a cycle, then the next 5, and then the next 7.
We can split a cycle up into a sequence of swaps that moves the first position down the line. For instance, the swaps $1\leftrightarrow 2, 2\leftrightarrow 3, 3\leftrightarrow 4$ in that order serve to ferry position 1 to the end of the four, while moving each other position one step left.
Doing this for each cycle of lengths $(4,5,7)$ gives us our exchange instructions. Then do the same for partner instructions to create cycles of $(3,13)$. Note that it doesn't matter which positions and dancers are chosen for the respective cycles.