Fix $a \geq 0$. Let $$F(n,a) = 2^{-n}\sum_{k=1}^{n-1} 2^{k} k^a.$$ Notice if $a=0$ this is just a finite geometric series. For $a>0$ we have $$\frac{F(n,a)}{n^a} \nearrow 1.$$ To see this one can write $$F(n,a) = \sum_{k=1}^{n-1} 2^{-k} (n-k)^a = \sum_{k=1}^{n-n^a} 2^{-k} (n-k)^a + \sum_{k=n-n^a+1}^{n-1} 2^{-k} (n-k)^a.$$ The first sum is asymptotic to $\sum_{k=1}^{n - n^a} 2^{-k} n^a \sim n^a$. The second sum is $O(n^{a^2})$ so vanishes when we divide by $n^a$.
I am interested in the rate of convergence in terms of $a$. More precisely, is there a function $f(n,a)$ so that $$1- \frac{F(n,a)}{n^a} \approx f(n,a).$$ I would only be interested in the leading order. Numerical approximations suggest that $f(n,a) \approx n^{-1}$ irregardless of $a$. I am somewhat distrustful of this, since it feels like the exponent should affect the convergence rate.
Probably not an answer but too long for a comment.
$$F(n,a) = 2^{-n}\sum_{k=1}^{n-1} 2^{k} k^a=2^{-n} \text{Li}_{-a}(2)-\Phi (2,-a,n)$$ where appear the polylogarithm function and the Lerch transcendent function.
Expanding for a few integer values of $a$ $$F(n,0)=1-2^{1-n}$$ $$F(n,1)=n-2+2^{1-n}$$ $$F(n,2)=n^2-4 n+6-3\times 2^{1-n}$$ $$F(n,3)=n^3-6 n^2+18 n-26+13\times 2^{1-n}$$ $$F(n,4)=n^4-8 n^3+36 n^2-104 n+150-75\times 2^{1-n}$$ $$F(n,5)=n^5-10 n^4+60 n^3-260 n^2+750 n-1082+541\times 2^{1-n}$$ $$F(n,6)=n^6-12 n^5+90 n^4-520 n^3+2250 n^2-6492 n+9366-4683\times 2^{1-n}$$ where we can notice some interesting patterns for the coefficients.
May be, this could help (I hope and wish).