Answer 2
There seems to be a general solution for this problem: for fractional "iterates" of any order, and possibly even for real "iterates". To show a solution for fractional "iterates" I'll simplify the formula in the question setting the exponent $p=1$ .
Let us assume an infinite set of coefficients $A=\{a_0,a_1,a_2,...\}$ to be determined, such that
$$ f(n) = a_0 \cdot n + a_1 \cdot (n-1) + ... + a_{n-1} \cdot 1 \tag 1$$
then to generalize this that let's introduce a notation for the "iteration" "height" $h$ as
$$ f°^{h+1}(n) = a_0 \cdot f°^h(n) + a_1 \cdot f°^h (n-1) + ... + a_{n-1} \cdot f°^h(1) \tag 2 $$
with $f°^h(0)=0$ for all $h$. (From this follows also $f°^h(1)=1$ for all $h$.) Let's identify notationally $f°^1(n)=f(n)$ from now.
solution for "half-iterates"
Now, if we want to make the second "iterate" equal to the formula for the sum-of-consecutive-$n$ ($s(n)=n+(n-1)+...+1$) we can expand the formula for the second iteration symbolically
using
$$ \begin{array} {}
f°^2(n) &=& a_0 \cdot f(n) + a_1 \cdot f (n-1) + ... + a_{n-1} \cdot f(1) \\
& = &a_0 \cdot (a_0 \cdot n + a_1 \cdot (n-1) + ... + a_{n-1} \cdot 1) \\
&& + a_1 \cdot (a_0 \cdot (n-1) + a_1 \cdot (n-2) + ... + a_{n-2} \cdot 1) \\
&& + a_2 \cdot (a_0 \cdot (n-2) + a_1 \cdot (n-3) + ... + a_{n-3} \cdot 1) \\
&& \vdots \\
&& + a_{n-1} \cdot (a_0 \cdot 1)
\end{array} \tag 3 $$
The coefficients can now be determined iteratively: we assume $n=1$ first and get
$$ \begin{array} {}
f°^2(1) &=& a_0 \cdot f(1) + a_1 \cdot f (0) + ... \\
& = &a_0 \cdot 1 + a_1 \cdot 0 + a_2 \cdot 0 + ... \\
&=& a_0
\end{array} \tag {4.0}$$
From comparision with $s(1)=1$ we conclude $a_0=1$.
Now we use $n=2$
$$ \begin{array} {}
f°^2(2) &=& 1\cdot f(2) + a_1 \cdot f (1) + 0 \\
& = & 1 \cdot (a_0 \cdot 2 + a_1 \cdot 1) + a_1 \cdot 1 \\
&=& 2 + 2 \cdot a_1 \\
\end{array} \tag {4.1a}$$
To equal this with $s(2)=3$ we solve
$$ 3 = 2 + 2 a_1 \\
1 = 2 a_1 \\
a_1 = 1/2 \tag {4.1b}
$$
This procedure repeated gives us the set of coefficients $A=\{1,1/2,3/8,10/32...\}$ with as many known coefficients as needed which I had already guessed by the "square-root of matrix G" ansatz.
For this specific function $f$ we can even find an independent formula for the set of coefficients, namely
$$ a_0=1 \qquad \qquad a_k = \binom{2k-1}{k} \cdot 2^{1-2k} \tag 5$$
which is suggested by the entry in OEIS.
So far we have used that algorithm with a simplified function $f(n)$ where we ignored exponents at $n, (n-1),...$ in $f°^1(n)=a_0 \cdot n + a_1 \cdot (n-1) + ... $. It is straightforward to show that the same coefficients $A$ occur, if we use similarly the generalized $s(n)$ (containing the same exponents) in the comparision-step.
general fractional "iterates"
Now what we've done to arrive at a set of coefficients $A$ to get some "half-iterate", we can analoguously determine such set for the "1/3-iterate" when we simply compare the values of $s(n)$ with $f°^3(n)$ - and the same can be made with any "$1/m$-iterate" for $m \in \mathbb N$ .
Here I show the beginning of the list of sets-of-coefficients $A_m$ for $m=1..8$
$\tag {6.1}$
a_0 a_1 a_2 ...
A_1: 1 1 1 1 1 1 1 1 1 1
A_2: 1 1/2 3/8 5/16 35/128 63/256 231/1024 429/2048 6435/32768 12155/65536
A_3: 1 1/3 2/9 14/81 35/243 91/729 728/6561 1976/19683 5434/59049 135850/1594323
A_4: 1 1/4 5/32 15/128 195/2048 663/8192 4641/65536 16575/262144 480675/8388608 1762475/33554432
A_5: 1 1/5 3/25 11/125 44/625 924/15625 4004/78125 17732/390625 79794/1953125 363506/9765625
A_6: 1 1/6 7/72 91/1296 1729/31104 8645/186624 267995/6718464 1416545/40310784 60911435/1934917632 2984660315/104485552128
A_7: 1 1/7 4/49 20/343 110/2401 638/16807 3828/117649 164604/5764801 1028775/40353607 6515575/282475249
A_8: 1 1/8 9/128 51/1024 1275/32768 8415/262144 115005/4194304 805035/33554432 45886995/2147483648 331406075/17179869184
A_9: 1 1/9 5/81 95/2187 665/19683 4921/177147 113183/4782969 889295/43046721 7114360/387420489 519348280/31381059609
A_10: 1 1/10 11/200 77/2000 2387/80000 97867/4000000 1663739/80000000 14498297/800000000 1029379087/64000000000 9264411783/640000000000
That coefficients can be normalized by multiplying them by powers of $m^2$. This gives the following decomposition:
$\tag {6.2}$
m A_m a_0 a_1 a_2 a_3 a_4 a_5 a_6
---+----+-----------------------------------------------------------------
1 A_1: 1 1/1^1 1/1^3 1/1^5 1/1^7 1/1^9 1/1^11
2 A_2: 1 1/2^1 3/2^3 10/2^5 35/2^7 126/2^9 462/2^11
3 A_3: 1 1/3^1 6/3^3 42/3^5 315/3^7 2457/3^9 19656/3^11
4 A_4: 1 1/4^1 10/4^3 120/4^5 1560/4^7 21216/4^9 297024/4^11
5 A_5: 1 1/5^1 15/5^3 275/5^5 5500/5^7 115500/5^9 2502500/5^11
6 A_6: 1 1/6^1 21/6^3 546/6^5 15561/6^7 466830/6^9 14471730/6^11
The sequences of numerators of the first couple of sets are also in the OEIS, for instance
entry in oeis first few coeffs short comment contributor
http://oeis.org/A000012 1,1,1,1
http://oeis.org/A001700 1,3,10,35,126
http://oeis.org/A034171 1,6,42,315,2457 "related to triple factorials" W.D.Lang
http://oeis.org/A034255 1,10,120,1560 "related to quartic factorials" W.D.Lang
http://oeis.org/A034687 1,15,275,5500 "related to quintic factorials" W.D.Lang
http://oeis.org/A034789 1,21,546,15561 "related to sextic factorials" W.D.Lang
alternative determination of coefficients $a_{m,k}$
That coefficients in row $m$ seem to fit polynomials in $1/m$ taken from the Stirlingnumbers $1$'st kind. Let $S1$ be the matrix of unsigned Stirling numbers $1$'st kind $S1 = [s_1(r,c)]_{r,c=0}^\infty$ such that
$\tag {7.1}$
1 . . . . .
0 1 . . . .
0 1 1 . . .
0 2 3 1 . . = top-left segment of S1
0 6 11 6 1 .
0 24 50 35 10 1
is the top left of $S1$ with row- and column-indexes beginning at zero, then the coefficients $a_{m,k}$ of the sets $A_m$ seem to be determined by
$$ a_{m,k} = \frac 1{m!} \cdot \sum_{c=0}^m {s1(m,c)\over m^c} \tag {7.2} $$
"real iterates"
In the comments at the OEIS-entries we find hints to generating-functions for those coefficients. Adapting that properly we find for the set $A_m$ the general generating function
$$ \mathcal{gf}(A_m) = (1-x)^{- 1/m} \tag {7.3}
$$
and because of that simple expression we might even generalize $m$ from natural numbers to real-numbers; for instance the set of coefficients $A_m$ such that "the $\pi$'th iterate" equals $s(n)$ are
$A_\pi =\{1, 0.318310, 0.209816, 0.162139, 0.134507, 0.116169, 0.102970, 0.0929424,...\}$