I found the lower bound $k = \left\lfloor \dfrac{n-1}{2} \right\rfloor$ with the sequence of jumps $n, \lfloor n/2 \rfloor, n-1, \lfloor n/2 \rfloor - 1, n-2, \lfloor n/2 \rfloor - 2, \ldots$ I also found a proof of the lower bound, but it's terribly complicated and I'm absolutely sure there is a simpler way. So please wait while I'm rephrasing my proof. The idea is to divide the jumps above in pairs and prove that each pair gives a net jump of exactly $k+1$ units, so the frog pretty much oscillates between $0$ and $k+1$. It remains to prove that it's the best one, as in it's also the upper bound.

Define the excess of a sequence of jumps to be the sum of the lengths of segments which do not cross over the entire range of $[1, k]$. Clearly, segments of length $1, \dots, k$ cannot leap over this range, so the excess is at least $1 + \dots + k$. Consider all leaps $i$ which go over the entire range $[1, k]$, and suppose each overshoots the range by some distance $d_i$ (i.e. $0$ and $k + 1$ get a $d$ of $0$, $-1$ and $k + 2$ get a distance of $1$, etc.) Then observe that since it overshot, it must start from the other side of the range $[1, k]$, so we have \[\sum d_i \le \sum_{i = k + 1} ^ n i - (k + 1) \]by assuming that all segments with length $\ge k + 1$ overshoot. However, $\sum d_i$ is an upper bound on the excess since each jump that doesn't leap over contributes to canceling out the excess. Thus, we get \[\sum_{i = k + 1} ^ n i - (k + 1) = \sum_{i = 1} ^ {n - k - 1} i \ge \sum_{i = 1} ^ k i.\]Hence, we require $n - k - 1 \ge k$, or $k \le \frac{n -1}{2}$. I claim $k = \left\lfloor \frac{n-1}{2} \right\rfloor$ is achievable. For $k = 2x$, simply take the sequence $2x, x, 2x-1, x-1, \dots, x+1, x$. For $k = 2x + 1$, simply take $2x + 1, x, 2x, x - 1, \dots, x + 2, 2, x + 1$. Grouping consecutive leaps together, we essentially makes leaps that exactly cross the gap. Thus, the answer is $\left\lfloor \dfrac{n-1}{2} \right\rfloor$.

I think you can establish the upper bound as follows as well - suppose $\lfloor(n-1)/2\rfloor+1$ were possible. Then at no time is the frog in $1,2,3,...,\lfloor(n-1)/2\rfloor+1$ but this means that at most negative the frog is at $\lfloor(n-1)/2\rfloor+2 - n = \lfloor(n-1)/2\rfloor - (n-2)$. Consider when the frog jumps by $\lfloor(n-1)/2\rfloor+1$. If the frog was on a positive number, that number is in $[\lfloor(n-1)/2\rfloor+2,n]$ so it ends in $[1,n-1 - \lfloor(n-1)/2\rfloor]$ which is within the forbidden region. If the frog was on nonpositive number, by the argument above it is in $[\lfloor(n-1)/2\rfloor - (n-2),0]$ so it ends within $[2*\lfloor(n-1)/2\rfloor - n-3,\lfloor(n-1)/2\rfloor+1]$, if n is odd then this is $[2,\lfloor(n-1)/2\rfloor+1]$, if n is even then this is $[1,\lfloor(n-1)/2\rfloor+1]$ but in both cases the frog is in the forbidden region