Edit: I now give the argument for general reductive $G$.
Let $G$ be a reductive algebraic group over an alg. closed field $k$ of char. 0. Fix a max
torus $T$ and write $X = X^*(T)$ for its group of characters. Write $R$ for the
subgroup of $X$ generated by the roots of $G$. Then the center $Z$
of $G$ is the diagonalizable subgroup of $T$ whose character group is $X/R$.
Claim: $G$ has a faithful irreducible representation if and only if the character
group $X/R$ of $Z$ is cyclic.
Note for semisimple $G$, the center $Z$ is finite. Since the characteristic of $k$ is 0, in this case the group of $k$-points of $Z$ is (non-canonically) isomorphic
to $X/R$. Thus $Z$ is cyclic if and only $X/R$ is cyclic.
In general, the condition that $X/R$ is cyclic means either that the group of points $Z(k)$ is finite cyclic, or that $Z$ is a 1 dimensional torus.
As to the proof, for $(\implies)$ see Boyarsky's comment following reb's answer.
For $(\Leftarrow)$ let me first treat the case where $G$ is almost simple; i.e. where the root system $\Phi$ of $G$ is irreducible. Supopse that the class of $\lambda \in X$ generates the cyclic group $X/R$. Since
the Weyl group acts on $X$ leaving $R$ invariant, the class of any $W$-conjugate of $\lambda$ is also a generator of $X/R$. Thus we may as well suppose $\lambda$ to be dominant
and non-0 [if $X=R$, take e.g. $\lambda$ to be a dominant root...]
Now the simple $G$-module $L=L(\lambda) = H^0(\lambda)$ with "highest weight $\lambda$"
will be faithful. To see this, note that since $\lambda \ne 0$, $L$ is not the trivial representation. Since $G$ is almost simple, the only proper normal subgroups
are contained in $Z$. Thus it suffices to observe that the action of $Z$ on the
$\lambda$ weight space of $L$ is faithful.
The general case is more-or-less the same, but with a bit more book-keeping.
Write the root system $\Phi$ of $G$ as a disjoint union $\Phi = \cup \Phi_i$
of its irreducible components. There is an isogeny
$$\pi:\prod_i G_{i,sc} \times T \to G$$
where $T$ is a torus and $G_{i,sc}$ is the simply connected almost simple group
with root system $\Phi_i$. Write $G_i$ for the image $\pi(G_{i,sc}) \subset G$.
The key fact is this:
a representation $\rho:G \to \operatorname{GL}(V)$ has
$\ker \rho \subset Z$ if and only if the restriction $\rho_{\mid G_i}$ is non-trivial
for each $i$.
Now, as before pick $\lambda \in X$ for which the coset of $\lambda$ generates
the assumed-to-be cyclic group $X/R$. After replacing $\lambda$ by a Weyl group
conjugate, we may suppose $\lambda$ to be dominant. After possibly repeatedly replacing
$\lambda$ by $\lambda + \alpha$ for dominant roots $\alpha$, we may suppose
that $\lambda$ has the following property:
$$(*) \quad \text{for each $i$, there is $\beta_i \in \Phi_i$ with $\langle \lambda,\beta_i^\vee \rangle \ne 0$}$$
Now let $L = L(\lambda)$ be the simple module with highest weight $\lambda$. Condition
$(*)$ implies that $G_i$ acts nontrivially on $L$ for each $i$, so by the "key fact",
the kernel of the representation of $G$ on $L$ lies in $Z$. but since $\lambda$
generates the group of characters of $Z$, the center $Z$ acts faithfully on the
$\lambda$ weight space of $L$.
You can mimic the standard finite group argument. Recall that, if $X$ and $Y$ are two finite-dimensional representations of $G$, with characters $\chi_X$ and $\chi_Y$, then $\dim \mathrm{Hom}_G(X,Y) = \int_G \overline{\chi_X} \otimes \chi_Y$, where the integral is with respect to Haar measure normalized so that $\int_G 1 =1$. The proof of this is exactly as in the finite case.
Now, let $W=1 \oplus V \oplus \overline{V}$. Your goal is to show that, for any nonzero representation $Y$, $\mathrm{Hom}(W^{\otimes N}, Y)$ is nonzero for $N$ sufficiently large. So you need to analyze $\int (1+\chi_V + \overline{\chi_V})^N \chi_W$ for $N$ large. Just as in the finite group case, we are going to split this into an integral near the identity, and an exponentially decaying term everywhere else.
I'm going to leave a lot of the analytic details to you, but here is the idea. Let $d=\dim V$. The function $f:=1+\chi_V + \overline{\chi_V}$ makes sense on the entire unitary group of $V$, of which $G$ is a subgroup. On a unitary matrix with eigenvalues $(e^{i \theta_1}, \cdots, e^{i \theta_d})$, we have $f=1+2 \sum \cos \theta_i$. In particular, for an element $g$ in the Lie algebra of $G$ near the identity, we have
$$f(e^g) = (2n+1) e^{-K(g) + O(|g|^4)}$$
where $K$ is $\mathrm{Tr}(g^* g) = \sum \theta_i^2$.
And, for $g$ near the identity, $\chi_X(e^g) = d + O(|g|)$. So the contribution to our integral near the identity can be approximated by
$$\int_{\mathfrak{g}} \left( (2n+1) e^{-K(g) + O(|g|^4)} \right)^N (d+ O(|g|) \approx (2n+1)^N d \int_{\mathfrak{g}} e^{-K(\sqrt{N} g)} = \frac{(2n+1)^N d}{N^{\dim G/2}} C$$
where $C$ is a certain Gaussian integral, and includes some sort of a factor concerning the determinant of the quadratic form $K$. You also need to work out what the Haar measure of $G$ turns into as a volume form on $\mathfrak{g}$ near $0$, I'll leave that to you as well.
For right now, the important point is that the growth rate is like $(2n+1)^N d$ divided by a polynomial factor. In the finite group case, that polynomial is a constant, which makes life easier, but we can live with a polynomial.
Now, look at the contribution from the rest of $G$. For any point in the unitary group $U(d)$, other than the identity, we have $|f| < 2n+1$. (To get equality, all the eigenvalues must be $1$ and, in the unitary group, that forces us to be at the identity.) So, if we break our integral into the integral over a small ball around the identity, plus an integral around everything else, the contribution of every thing else will be $O(a^N)$ with $a<2n+1$.
So, just as in the finite group case, the exponential with the larger base wins, and the polynomial in the denominator is too small to effect the argument. Joel and I discussed a similar, but harder, argument over at the Secret Blogging Seminar.
Best Answer
A famous theorem is that this is true if and only if $G$ is a Lie group.