Strömbergsson's footnote refers to $\sum_{|n|\leq N}|c_n|^2$, not $\sum_{|n|\leq N}|n||c_n|^2$. In fact his $c_n$ is Iwaniec's $|n|^{1/2}\hat f_\mathfrak{a}(n)$. Normalization is a serious difficulty in the subject, one must be careful.
Strömbergsson points out that Iwaniec's proof is incorrect, but the result is fine. To fix the given line in Iwaniec's proof, integrate from $|s|/4$ instead of $|s|$, then the claimed bound is all right. This you can see from the discussion below (4.13) in Strömbergsson's paper: already the integral from $|s|/4$ to $|s|/2$ is large. Note that Strömbergsson talks about $s=1/2+iR$ (i.e. the case $\Re s=1/2$), but for the remaining $s$ (i.e. $1/2\leq s\leq 1$) Iwaniec's bound follows by compactness.
I don't know (from the top of my head) of any better uniform estimate than the one displayed. In special ranges or for Hecke eigenforms one has better estimates, but there are many open questions (e.g. how to give a good lower bound when $N$ is small compared to the various parameters of the form).
Added. This is a response to the OP's comment below. If $\lambda(n)$ denotes the Hecke eigenvalues of a primitive cusp form, then one can derive an asymptotic formula for $\sum_{n\leq N} |\lambda(n)|^2$ or a smooth version of it by applying standard Mellin transform techniques for the Dirichlet series $\sum_{n=1}^\infty |\lambda(n)|^2/n^s=L(s,f\otimes\tilde f)/\zeta(2s)$. The main term comes from the pole at $s=1$, while the error term depends on bounds for the Rankin-Selberg $L$-function $L(s,f\otimes\tilde f)$ to the left of $s=1$. The main term involves the residue at $s=1$, for which very good bounds are available by Iwaniec (Acta Arith. 56 (1990), 65–82) and by Hoffstein-Lockhart (Ann. of Math. 140 (1994), 161–181). For the error term one can use the standard convexity bound for $L(s,f\otimes\tilde f)$, and hopefully a subconvex bound will also be available later in the future. Rankin in his original paper (Proc. Cambridge Philos. Soc. 35 (1939), 351–372) established $\sum_{n\leq N} |\lambda(n)|^2=cN+O(N^{3/5})$, where the error term is still the best one and follows from a general result of Landau available here.
