If $\mathbf{v}\in A$ is in the kernel of $\phi_i$, then for all $j\leq i$ you have $\phi_j(\mathbf{v}) = f_{ij}\phi_j(\mathbf{v}) = f_{ij}(\mathbf{0}) = \mathbf{0}$ (where $f_{ij}\colon A_i\to A_j$ is the map obtained by composing the $f_k$). Therefore, $\mathrm{ker}(\phi_i)\subseteq \mathrm{ker}(\phi_j)$ for all $i\geq j$. Hence you have a descending sequence
\begin{equation*}
\cdots \subseteq \mathrm{ker}(\phi_{i+1})\subseteq \mathrm{ker}(\phi_i)\subseteq \cdots \subseteq \mathrm{ker}(\phi_0).
\end{equation*}
Since $\dim(A)\lt\infty$, the sequence must stabilize, so there exists $n_0$ such that for all $n\geq n_0$, $\mathrm{ker}(\phi_n)=\mathrm{ker}(\phi_{n_0})$. Let $\mathbf{N}$ be that subspace of $A$.
If $\mathbf{v}\in\mathbf{N}$, then $\phi_i(\mathbf{v})=\mathbf{0}$ for all $i$, hence $\phi(\mathbf{v})=\mathbf{0}$, so $\mathbf{N}\subseteq \mathrm{ker}(\phi)$. Conversely, anything in $\mathrm{ker}(\phi)$ lies in the intersection of all kernels of the $\phi_i$. Thus, $\mathbf{N}=\mathrm{ker}(\phi)$.
In conclusion, $\phi$ is one to one if and only if the $\phi_i$ are eventually one-to-one.
For surjection, you are going to be out of luck. Consider the system with $A_i = \mathbf{F}^2$ for all $i$, $f_i\colon\mathbf{F}^2\to\mathbf{F}^2$ given by $f_i(a,b)=(a,0)$. I claim that the inverse limit of this system is $\mathbf{V}=\mathbf{F}$ with structure maps $\mathfrak{f}_i\colon\mathbf{V}\to\mathbf{F}^2$ given by $\mathfrak{f}(a) = (a,0)$. First, note that this is a consistent system of maps from $\mathbf{V}$ to the $A_i$. Next, suppose that $\mathbf{W}$ is any vector space and that you have maps $g_i\colon\mathbf{W}\to\mathbf{F}^2$ for all $i$, with the property that for all $i\geq j$, $g_j(\mathbf{w}) = f_{ij}(g_i(\mathbf{w}))$. If $g_i(\mathbf{w}) = (\mathbf{w}_{1i},\mathbf{w}_{2i})$, then this means that $g_j(\mathbf{w}) = (\mathbf{w}_{1i}, 0)$. In particular, for all $j$ we have that $\mathrm{Im}(g_j)$ is contained in the subspace $\{(x,0)\mid x\in\mathbf{F}\}$ of $A_j$, and since the $f_i$ are isomorphisms when restricted to those subspaces, then it follows that for each $\mathbf{w}\in\mathbf{W}$ there is an $a_{\mathbf{w}}\in\mathbf{F}$ such that $g_j(\mathbf{w}) = (a_{\mathbf{w}},0)$ for all $j$. Hence, we have a map $\mathfrak{g}\colon\mathbf{W}\to \mathbf{V}$ with $\mathfrak{f}_j\circ\mathfrak{g}=g_j$ for all $j$, namely $\mathfrak{g}(\mathbf{w}) = a_{\mathbf{w}}$, and the map is clearly unique. Thus, $(\mathbf{V},\mathfrak{f}_i)$ is the inverse limit of the system. Then $\mathbf{V}$ itself with the structure maps gives a counterexample: the induced map to the inverse limit is onto (in fact, it is the identity), but the structure maps are never onto, so it is false that $\phi_i$ is onto for sufficiently large $i$.
Added: However, if your structure maps $f_i$ are eventually surjective, then your conclusion will follow. To see this, assume that $f_i$ is surjective for all $i\geq N$; I claim that the inverse limit maps $\mathfrak{f}_j\colon\lim\limits_{\leftarrow}A_i\to A_j$ are surjective for all $j\geq N$. Fix $j\geq N$, and let $\mathbf{v}\in A_j$. Let $\mathbf{v}_{j+1}\in A_{j+1}$ be any preimage of $\mathbf{v}$, and inductively define $\mathbf{v}_{j+k+1}$ to be any preimage of $\mathbf{v}_{j+k}$. Then define maps $g_i\colon\mathbf{F}\to A_i$ by letting $g_i(1) = \mathbf{v}_{j+m}$ if $i=j+m\geq N$, and $g_i(1)=f_{ji}(\mathbf{v})$ if $i\lt N$. This induces a map $\mathfrak{g}\colon\mathbf{F}\to\lim\limits_{\leftarrow}A_i$, and $g_i = \mathfrak{f}_i\circ\mathfrak{g}$. In particular, $\mathbf{v}\in\mathrm{Im}(\mathfrak{f}_j)$. Thus, $\mathfrak{f}_j$ is onto for all $j\geq N$. Then, if you have $\phi\colon A\to A_i$ as in your statement surjective, then $\phi_j = \mathfrak{f}_j\circ\phi\colon A\to A_j$ is a composition of surjective functions for alL $j\geq N$, so the $\phi_j$ are eventually surjective, as desired.
Added 2: To see how I came up with the example, notice that if you let $B_{i-1} = \mathrm{Im}(f_i)$, then you get an inverse system with the $B_i$ instead of the $A_i$; the inverse limit of this system is isomorphic to the inverse limit of the original system (try to prove this). This is because a consistent system of homomorphism $\phi_i\colon A\to A_i$ has to have $\mathrm{Im}(\phi_i)\subseteq B_i$. So what really matters is what is going on in the images of the structure maps $f_i$, hence surjectivity into the limit depends only on how your maps $f_i$ relate to the $B_i$, not how they relate to the $A_i$. That is why if the maps $f_i$ are eventually surjective you do get the implication.
(1). There are endomorphisms $T$ with $\ker(T)=\{0\}$ which are not surjective.
(2). Not in every case a linear form $\phi$ is representable by a vector $v$ in presence of a scalar product, i.e., there doesn't exist a vector $v$ that $\phi(.)=\langle v,.\rangle$.
(3). Not all linear mappings are continuos.
(4). You can equip a vector space with two different norms such that the unit ball in respect to the first norm in unbounded in respect to the second.
It's just a brand new world.
Best Answer
For simplicity, all the vector spaces in the following are over $\mathbb{C}$, or some complete field.
All norms on a finite dimensional vector space are equivalent. This is not true for infinite dimensional vector spaces over (consider $L^p$ norms). I believe this comes from the fact that the unit ball is compact for a finite dimensional normed linear spaces (NLS), but not in infinite dimensional NLS.
The weak topology on a finite dimensional vector space is equivalent to the norm topology. This is always false for infinite dimensional vector spaces. More generally, there are many topologies of interest on an infinite dimensional vector space, but just one of interest on a finite dimensional space (from a linear algebra/functional analysis perspective).
There is a nontrivial translation invariant measure for finite dimensional vector spaces (say over $\mathbb{C} $ or $ \mathbb{R}$, the Lebesgue measure). This is not true for an infinite dimensional Hilbert space (the unit ball has infinitely many disjoint translates of a ball of radius $\sqrt{2}/4$).
This is a property of bounded operators on a NLS: The spectrum of a linear map $T:V \to V$ (the set of $\lambda\in \mathbb{C}$ such that $T-\lambda I$ is not invertible) consists precisely of the eigenvalues of $T$. However, if $V$ is infinite dimensional, then $T-\lambda I$ may not be invertible even if $\lambda$ is not an eigenvalue.