$\cos(x)$ and $\arccos(x)$ couple limit
A more convenient way to state the sequence is: $$ \frac{\sum_{k=1}^n\cos\frac{1}{k}\arccos\frac{1}{n-k+1}}{n} $$
Note that for $0\leq x\leq 1$ we have $1-x\leq\cos x\leq 1$ and $\frac{\pi}{2}-\frac{\pi}{2}x\leq\arccos x\leq \frac{\pi}{2}-x$. Therefore, we have $$ \tfrac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})\leq\cos\frac{1}{k}\arccos\frac{1}{n-k+1}\leq\frac{\pi}{2}-\frac{1}{n-k+1}. $$ Thus, a lower bound for the limit is given by the sequence $$ \frac{\sum_{k=1}^n\frac{\pi}{2}(1-\tfrac{1}{k})(1-\tfrac{1}{n-k+1})}{n}=\frac{\pi}{2}-\frac{\sum_{k=0}^n\frac{(n-k+1)+k-1}{k(n-k+1)}}{n}=\frac{\pi}{2}-\sum_{k=1}^n\frac{1}{k(n-k+1)} $$ The terms in the latter sum can be rewritten to $$ \frac{1}{k(n+1)}+\frac{1}{(n-k+1)(n+1)} $$ so we get the sums $$ \frac{1}{n+1}\sum_{k=1}^n\frac{1}{k}\qquad\text{and}\qquad\frac{1}{n+1}\sum_{k=1}^n\frac{1}{n-k+1}. $$ Sacha indicated a proof in the comments that these sums tend to zero. Similarly, for the upper bound we have $$ \frac{\sum_{k=1}^n\frac{\pi}{2}-\tfrac{1}{n-k+1}}{n}=\frac{\pi}{2}-\frac{\sum_{k=1}^n\frac{1}{n-k+1}}{n}\to\frac{\pi}{2} $$ This gives an upper bound of $\frac{\pi}{2}$. This finishes the proof that the limit is $\frac{\pi}{2}$.
Note: Another proof that $\frac{1}{n}\sum_{k=1}^n\frac{1}{k}\to 0$. Using Cauchy-Schwarz, we see that $$ \sum_{k=1}^n\frac{1}{k}\leq \sqrt{n}\sqrt{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}} $$ Therefore we get $$ \frac{1}{n}\sum_{k=1}^n\frac{1}{k}\leq\sqrt{\frac{\textstyle\sum_{k=1}^n\tfrac{1}{k^2}}{n}} $$ In the square root on the right hand side, the numerator converges, so the whole tends to zero.
What follows is a little hand-wavy, and I wish I had more rigorous demonstration, but the post is too big for a comment.
$$ \begin{eqnarray} \frac{1}{n} \sum_{k=1}^n \cos\left(\frac{1}{k}\right) \arccos\left( \frac{1}{n+1-k} \right) &=& \frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right) \end{eqnarray} $$ We can now split the sum in two parts, $1 \leqslant k \leqslant \lfloor\frac{n}{2}\rfloor$ and $\lfloor\frac{n}{2}\rfloor < k \leqslant n$. In each of these parts, either $\sin$, or $\arcsin$ will be small, and in the limiting value will be $\frac{\pi}{2}$:
In[28]:= Table[
N[1/n Sum[Cos[1/k] ArcCos[1/(n - k + 1)], {k, 1, n}],
50], {n, {5000, 10000, 100000}}] // N
Out[28]= {1.56861, 1.56963, 1.57066}
More rigorously: $$ \frac{1}{n} \sum_{k=1}^n \left( 1 - 2 \sin^2\left(\frac{1}{2 k}\right) \right) \left( \frac{\pi}{2} - \arcsin\left( \frac{1}{n+1-k} \right) \right) = \frac{\pi}{2} - \frac{\pi}{n} \sum_{k=1}^n \sin^2 \frac{1}{k} - \frac{1}{n} \sum_{k=1}^n \arcsin\frac{1}{k} + \frac{2}{n} \sum_{k=1}^n \sin^2\left( \frac{1}{k}\right) \arcsin\left(\frac{1}{n+1-k}\right) $$ Now: $$ 0 \leqslant \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \sin^2\left(\frac{1}{k}\right) \leqslant \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{k^2} = 0 $$ $$ 0 \leqslant \lim_{n\to \infty} \frac{1}{n} \sum_{k=1}^n \arcsin\left(\frac{1}{k}\right) \leqslant \lim_{n \to \infty} \frac{\pi}{2n} \sum_{k=1}^n \frac{1}{k} = \lim_{n \to \infty} \frac{\pi}{2 n} \ln(n) = 0 $$ $$ 0 \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \sin^2 \left( \frac{1}{k} \right) \arcsin\left(\frac{1}{n+1-k} \right) \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} \frac{1}{n+1-k} \leqslant \lim_{n \to \infty} \frac{2}{n} \sum_{k=1}^n \frac{1}{k^2} = 0 $$